State if the statement is True or False: The maximum value of $2x^3-9x^2-24x-20$ is $-7$.












0















State if the statement is True or False:
The maximum value of $2x^3-9x^2-24x-20$ is $-7$.




Let $f(x) = 2x^3-9x^2-24x-20$.



If we go by the derivative test:
$$f'(x) = 6x^2-18x-24 & f'(x) = 0 implies x=4,-1$$



At $x=-1$ we get $f(x)=-7$ and at $x=4$ we get $f(x) = -132$, so we have maximum value $-7$ by this method.



But this is a polynomial function, its value tends to infinity as $x to infty$.



So what can be said about the truth of the statement?






real-analysis analysis functions maxima-minima







share|cite|improve this question






















  • Every local maximum $x$ of $f$ has $f'(x) = 0$, since $frac{f(x + h) - f(x)}{h}$ is negative for small $h > 0$ and positive for small $h < 0$; since $f$ is differentiable at $x$, the limit must equal $0$. That's necessary but not sufficient; as you note, $f$ is clearly unbounded.
    – anomaly
    Nov 22 '18 at 2:44










  • $f'(x) = 0 at local minimums also. Look at the dominating first term for the global max and mins.
    – Phil H
    Nov 22 '18 at 3:03






  • 4




    When they say "maximum" or "minimum" without qualification, they are generally referring to global phenomena, not local phenomena. So the statement is false.
    – Deepak
    Nov 22 '18 at 3:19


















0















State if the statement is True or False:
The maximum value of $2x^3-9x^2-24x-20$ is $-7$.




Let $f(x) = 2x^3-9x^2-24x-20$.



If we go by the derivative test:
$$f'(x) = 6x^2-18x-24 & f'(x) = 0 implies x=4,-1$$



At $x=-1$ we get $f(x)=-7$ and at $x=4$ we get $f(x) = -132$, so we have maximum value $-7$ by this method.



But this is a polynomial function, its value tends to infinity as $x to infty$.



So what can be said about the truth of the statement?






real-analysis analysis functions maxima-minima







share|cite|improve this question






















  • Every local maximum $x$ of $f$ has $f'(x) = 0$, since $frac{f(x + h) - f(x)}{h}$ is negative for small $h > 0$ and positive for small $h < 0$; since $f$ is differentiable at $x$, the limit must equal $0$. That's necessary but not sufficient; as you note, $f$ is clearly unbounded.
    – anomaly
    Nov 22 '18 at 2:44










  • $f'(x) = 0 at local minimums also. Look at the dominating first term for the global max and mins.
    – Phil H
    Nov 22 '18 at 3:03






  • 4




    When they say "maximum" or "minimum" without qualification, they are generally referring to global phenomena, not local phenomena. So the statement is false.
    – Deepak
    Nov 22 '18 at 3:19
















0












0








0


1






State if the statement is True or False:
The maximum value of $2x^3-9x^2-24x-20$ is $-7$.




Let $f(x) = 2x^3-9x^2-24x-20$.



If we go by the derivative test:
$$f'(x) = 6x^2-18x-24 & f'(x) = 0 implies x=4,-1$$



At $x=-1$ we get $f(x)=-7$ and at $x=4$ we get $f(x) = -132$, so we have maximum value $-7$ by this method.



But this is a polynomial function, its value tends to infinity as $x to infty$.



So what can be said about the truth of the statement?






real-analysis analysis functions maxima-minima







share|cite|improve this question














State if the statement is True or False:
The maximum value of $2x^3-9x^2-24x-20$ is $-7$.




Let $f(x) = 2x^3-9x^2-24x-20$.



If we go by the derivative test:
$$f'(x) = 6x^2-18x-24 & f'(x) = 0 implies x=4,-1$$



At $x=-1$ we get $f(x)=-7$ and at $x=4$ we get $f(x) = -132$, so we have maximum value $-7$ by this method.



But this is a polynomial function, its value tends to infinity as $x to infty$.



So what can be said about the truth of the statement?






real-analysis analysis functions maxima-minima




real-analysis analysis functions maxima-minima






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 22 '18 at 2:41









user8795

5,61961947




5,61961947












  • Every local maximum $x$ of $f$ has $f'(x) = 0$, since $frac{f(x + h) - f(x)}{h}$ is negative for small $h > 0$ and positive for small $h < 0$; since $f$ is differentiable at $x$, the limit must equal $0$. That's necessary but not sufficient; as you note, $f$ is clearly unbounded.
    – anomaly
    Nov 22 '18 at 2:44










  • $f'(x) = 0 at local minimums also. Look at the dominating first term for the global max and mins.
    – Phil H
    Nov 22 '18 at 3:03






  • 4




    When they say "maximum" or "minimum" without qualification, they are generally referring to global phenomena, not local phenomena. So the statement is false.
    – Deepak
    Nov 22 '18 at 3:19




















  • Every local maximum $x$ of $f$ has $f'(x) = 0$, since $frac{f(x + h) - f(x)}{h}$ is negative for small $h > 0$ and positive for small $h < 0$; since $f$ is differentiable at $x$, the limit must equal $0$. That's necessary but not sufficient; as you note, $f$ is clearly unbounded.
    – anomaly
    Nov 22 '18 at 2:44










  • $f'(x) = 0 at local minimums also. Look at the dominating first term for the global max and mins.
    – Phil H
    Nov 22 '18 at 3:03






  • 4




    When they say "maximum" or "minimum" without qualification, they are generally referring to global phenomena, not local phenomena. So the statement is false.
    – Deepak
    Nov 22 '18 at 3:19


















Every local maximum $x$ of $f$ has $f'(x) = 0$, since $frac{f(x + h) - f(x)}{h}$ is negative for small $h > 0$ and positive for small $h < 0$; since $f$ is differentiable at $x$, the limit must equal $0$. That's necessary but not sufficient; as you note, $f$ is clearly unbounded.
– anomaly
Nov 22 '18 at 2:44




Every local maximum $x$ of $f$ has $f'(x) = 0$, since $frac{f(x + h) - f(x)}{h}$ is negative for small $h > 0$ and positive for small $h < 0$; since $f$ is differentiable at $x$, the limit must equal $0$. That's necessary but not sufficient; as you note, $f$ is clearly unbounded.
– anomaly
Nov 22 '18 at 2:44












$f'(x) = 0 at local minimums also. Look at the dominating first term for the global max and mins.
– Phil H
Nov 22 '18 at 3:03




$f'(x) = 0 at local minimums also. Look at the dominating first term for the global max and mins.
– Phil H
Nov 22 '18 at 3:03




4




4




When they say "maximum" or "minimum" without qualification, they are generally referring to global phenomena, not local phenomena. So the statement is false.
– Deepak
Nov 22 '18 at 3:19






When they say "maximum" or "minimum" without qualification, they are generally referring to global phenomena, not local phenomena. So the statement is false.
– Deepak
Nov 22 '18 at 3:19












5 Answers
5






active

oldest

votes


















1














Hint: Let $x=10^{20}$. That should do it.






share|cite|improve this answer

















  • 1




    x=10 will also give positive value!
    – user8795
    Nov 22 '18 at 2:49










  • Well, there's your answer, @user8795.
    – Shaun
    Nov 22 '18 at 2:51










  • I think @anomaly's comment should be an answer and my answer should have been a comment. If you're looking for a quick, simple answer, though, mine will do the job if you recall the definition of what a maximum is.
    – Shaun
    Nov 22 '18 at 2:55



















0














$f^{primeprime}(-1)=12(-1)-18=-30<0$. Thus $f$ has a local maximum at $-1$ and the local maximum value is $f(-1)=-7$. But it does not mean $f$ can not assume any value greater than $-7$. In fact the function $f$ is clearly unbounded. Here $f^{prime}(x)>0$ for all $xin (-infty,-1)$ and for all $xin (4,infty)$ and $f^{prime}(x)<0$ for all $xin (-1,4)$. Thus $f$ is strictly increasing in $(-infty,-1)cup (4,infty)$ and is strictly decreasing in $(-1,4)$. At $-1$ it has a local maximum and at $4$ it has a local minimum.






share|cite|improve this answer





















  • so the conclusion is false?
    – user8795
    Nov 22 '18 at 3:15








  • 2




    Unless the word "local" is mentioned, it is certainly false.
    – Anupam
    Nov 22 '18 at 3:17



















0














Assume the opposite:



$$2x^3-9x^2-24x-20>-7$$
$$to 2x^3-9x^2-24x-13>0$$



$$to (2x-13)(x+1)^2>0$$
Pretty blatant from this.






share|cite|improve this answer





















  • Your use of implication (" $to$ ") is not enough. One must be able to go from your last inequality to your first. Indeed, this is possible, but you haven't made that clear.
    – Shaun
    Nov 23 '18 at 3:47



















0














For large enough $x$, the polynomial grows unboundedly, therefore no finite maximum exists for the function. The points then you found, are just local maximum or minimum not global






share|cite|improve this answer





























    0














    The differentiation & setting of the derivative to zero yields stationary points. A range-of-x ought to have been specified for this question, as without it it is ill-posed.



    To say that the function has a maximum value of ∞ at x=∞ is a scrambling of the meaning of what's going-on, which is that the function increases without limit as x increases without limit, & therefore does not have a maximum value. If a range of x had been given, you would compare the -7 gotten as a local maximum to the value taken by the function at the upper end of the range of x (you need not consider the lower end of the range of x as the function is decreasing without limit in that direction) ... and whichever were the greater would be the answer to the question.



    A fussy little point though: but it's important in many kinds of problem: the range would have to be a closed range - ie of the form $$aleq xleq b ,$$ often denoted $$xin[a,b], $$ rather than what is called an open range $$a< x< b ,$$ often denoted $$xin(a,b) ;$$ as if it were the latter that were specified, you would technically still have the problem of the non-existence of a maximum value of the function if the upper limit of the range were such that it could exceed -7 for x still less than it.






    share|cite|improve this answer























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      5 Answers
      5






      active

      oldest

      votes








      5 Answers
      5






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1














      Hint: Let $x=10^{20}$. That should do it.






      share|cite|improve this answer

















      • 1




        x=10 will also give positive value!
        – user8795
        Nov 22 '18 at 2:49










      • Well, there's your answer, @user8795.
        – Shaun
        Nov 22 '18 at 2:51










      • I think @anomaly's comment should be an answer and my answer should have been a comment. If you're looking for a quick, simple answer, though, mine will do the job if you recall the definition of what a maximum is.
        – Shaun
        Nov 22 '18 at 2:55
















      1














      Hint: Let $x=10^{20}$. That should do it.






      share|cite|improve this answer

















      • 1




        x=10 will also give positive value!
        – user8795
        Nov 22 '18 at 2:49










      • Well, there's your answer, @user8795.
        – Shaun
        Nov 22 '18 at 2:51










      • I think @anomaly's comment should be an answer and my answer should have been a comment. If you're looking for a quick, simple answer, though, mine will do the job if you recall the definition of what a maximum is.
        – Shaun
        Nov 22 '18 at 2:55














      1












      1








      1






      Hint: Let $x=10^{20}$. That should do it.






      share|cite|improve this answer












      Hint: Let $x=10^{20}$. That should do it.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Nov 22 '18 at 2:45









      Shaun

      8,810113680




      8,810113680








      • 1




        x=10 will also give positive value!
        – user8795
        Nov 22 '18 at 2:49










      • Well, there's your answer, @user8795.
        – Shaun
        Nov 22 '18 at 2:51










      • I think @anomaly's comment should be an answer and my answer should have been a comment. If you're looking for a quick, simple answer, though, mine will do the job if you recall the definition of what a maximum is.
        – Shaun
        Nov 22 '18 at 2:55














      • 1




        x=10 will also give positive value!
        – user8795
        Nov 22 '18 at 2:49










      • Well, there's your answer, @user8795.
        – Shaun
        Nov 22 '18 at 2:51










      • I think @anomaly's comment should be an answer and my answer should have been a comment. If you're looking for a quick, simple answer, though, mine will do the job if you recall the definition of what a maximum is.
        – Shaun
        Nov 22 '18 at 2:55








      1




      1




      x=10 will also give positive value!
      – user8795
      Nov 22 '18 at 2:49




      x=10 will also give positive value!
      – user8795
      Nov 22 '18 at 2:49












      Well, there's your answer, @user8795.
      – Shaun
      Nov 22 '18 at 2:51




      Well, there's your answer, @user8795.
      – Shaun
      Nov 22 '18 at 2:51












      I think @anomaly's comment should be an answer and my answer should have been a comment. If you're looking for a quick, simple answer, though, mine will do the job if you recall the definition of what a maximum is.
      – Shaun
      Nov 22 '18 at 2:55




      I think @anomaly's comment should be an answer and my answer should have been a comment. If you're looking for a quick, simple answer, though, mine will do the job if you recall the definition of what a maximum is.
      – Shaun
      Nov 22 '18 at 2:55











      0














      $f^{primeprime}(-1)=12(-1)-18=-30<0$. Thus $f$ has a local maximum at $-1$ and the local maximum value is $f(-1)=-7$. But it does not mean $f$ can not assume any value greater than $-7$. In fact the function $f$ is clearly unbounded. Here $f^{prime}(x)>0$ for all $xin (-infty,-1)$ and for all $xin (4,infty)$ and $f^{prime}(x)<0$ for all $xin (-1,4)$. Thus $f$ is strictly increasing in $(-infty,-1)cup (4,infty)$ and is strictly decreasing in $(-1,4)$. At $-1$ it has a local maximum and at $4$ it has a local minimum.






      share|cite|improve this answer





















      • so the conclusion is false?
        – user8795
        Nov 22 '18 at 3:15








      • 2




        Unless the word "local" is mentioned, it is certainly false.
        – Anupam
        Nov 22 '18 at 3:17
















      0














      $f^{primeprime}(-1)=12(-1)-18=-30<0$. Thus $f$ has a local maximum at $-1$ and the local maximum value is $f(-1)=-7$. But it does not mean $f$ can not assume any value greater than $-7$. In fact the function $f$ is clearly unbounded. Here $f^{prime}(x)>0$ for all $xin (-infty,-1)$ and for all $xin (4,infty)$ and $f^{prime}(x)<0$ for all $xin (-1,4)$. Thus $f$ is strictly increasing in $(-infty,-1)cup (4,infty)$ and is strictly decreasing in $(-1,4)$. At $-1$ it has a local maximum and at $4$ it has a local minimum.






      share|cite|improve this answer





















      • so the conclusion is false?
        – user8795
        Nov 22 '18 at 3:15








      • 2




        Unless the word "local" is mentioned, it is certainly false.
        – Anupam
        Nov 22 '18 at 3:17














      0












      0








      0






      $f^{primeprime}(-1)=12(-1)-18=-30<0$. Thus $f$ has a local maximum at $-1$ and the local maximum value is $f(-1)=-7$. But it does not mean $f$ can not assume any value greater than $-7$. In fact the function $f$ is clearly unbounded. Here $f^{prime}(x)>0$ for all $xin (-infty,-1)$ and for all $xin (4,infty)$ and $f^{prime}(x)<0$ for all $xin (-1,4)$. Thus $f$ is strictly increasing in $(-infty,-1)cup (4,infty)$ and is strictly decreasing in $(-1,4)$. At $-1$ it has a local maximum and at $4$ it has a local minimum.






      share|cite|improve this answer












      $f^{primeprime}(-1)=12(-1)-18=-30<0$. Thus $f$ has a local maximum at $-1$ and the local maximum value is $f(-1)=-7$. But it does not mean $f$ can not assume any value greater than $-7$. In fact the function $f$ is clearly unbounded. Here $f^{prime}(x)>0$ for all $xin (-infty,-1)$ and for all $xin (4,infty)$ and $f^{prime}(x)<0$ for all $xin (-1,4)$. Thus $f$ is strictly increasing in $(-infty,-1)cup (4,infty)$ and is strictly decreasing in $(-1,4)$. At $-1$ it has a local maximum and at $4$ it has a local minimum.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Nov 22 '18 at 3:14









      Anupam

      2,3741823




      2,3741823












      • so the conclusion is false?
        – user8795
        Nov 22 '18 at 3:15








      • 2




        Unless the word "local" is mentioned, it is certainly false.
        – Anupam
        Nov 22 '18 at 3:17


















      • so the conclusion is false?
        – user8795
        Nov 22 '18 at 3:15








      • 2




        Unless the word "local" is mentioned, it is certainly false.
        – Anupam
        Nov 22 '18 at 3:17
















      so the conclusion is false?
      – user8795
      Nov 22 '18 at 3:15






      so the conclusion is false?
      – user8795
      Nov 22 '18 at 3:15






      2




      2




      Unless the word "local" is mentioned, it is certainly false.
      – Anupam
      Nov 22 '18 at 3:17




      Unless the word "local" is mentioned, it is certainly false.
      – Anupam
      Nov 22 '18 at 3:17











      0














      Assume the opposite:



      $$2x^3-9x^2-24x-20>-7$$
      $$to 2x^3-9x^2-24x-13>0$$



      $$to (2x-13)(x+1)^2>0$$
      Pretty blatant from this.






      share|cite|improve this answer





















      • Your use of implication (" $to$ ") is not enough. One must be able to go from your last inequality to your first. Indeed, this is possible, but you haven't made that clear.
        – Shaun
        Nov 23 '18 at 3:47
















      0














      Assume the opposite:



      $$2x^3-9x^2-24x-20>-7$$
      $$to 2x^3-9x^2-24x-13>0$$



      $$to (2x-13)(x+1)^2>0$$
      Pretty blatant from this.






      share|cite|improve this answer





















      • Your use of implication (" $to$ ") is not enough. One must be able to go from your last inequality to your first. Indeed, this is possible, but you haven't made that clear.
        – Shaun
        Nov 23 '18 at 3:47














      0












      0








      0






      Assume the opposite:



      $$2x^3-9x^2-24x-20>-7$$
      $$to 2x^3-9x^2-24x-13>0$$



      $$to (2x-13)(x+1)^2>0$$
      Pretty blatant from this.






      share|cite|improve this answer












      Assume the opposite:



      $$2x^3-9x^2-24x-20>-7$$
      $$to 2x^3-9x^2-24x-13>0$$



      $$to (2x-13)(x+1)^2>0$$
      Pretty blatant from this.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Nov 22 '18 at 5:59









      Rhys Hughes

      5,0091427




      5,0091427












      • Your use of implication (" $to$ ") is not enough. One must be able to go from your last inequality to your first. Indeed, this is possible, but you haven't made that clear.
        – Shaun
        Nov 23 '18 at 3:47


















      • Your use of implication (" $to$ ") is not enough. One must be able to go from your last inequality to your first. Indeed, this is possible, but you haven't made that clear.
        – Shaun
        Nov 23 '18 at 3:47
















      Your use of implication (" $to$ ") is not enough. One must be able to go from your last inequality to your first. Indeed, this is possible, but you haven't made that clear.
      – Shaun
      Nov 23 '18 at 3:47




      Your use of implication (" $to$ ") is not enough. One must be able to go from your last inequality to your first. Indeed, this is possible, but you haven't made that clear.
      – Shaun
      Nov 23 '18 at 3:47











      0














      For large enough $x$, the polynomial grows unboundedly, therefore no finite maximum exists for the function. The points then you found, are just local maximum or minimum not global






      share|cite|improve this answer


























        0














        For large enough $x$, the polynomial grows unboundedly, therefore no finite maximum exists for the function. The points then you found, are just local maximum or minimum not global






        share|cite|improve this answer
























          0












          0








          0






          For large enough $x$, the polynomial grows unboundedly, therefore no finite maximum exists for the function. The points then you found, are just local maximum or minimum not global






          share|cite|improve this answer












          For large enough $x$, the polynomial grows unboundedly, therefore no finite maximum exists for the function. The points then you found, are just local maximum or minimum not global







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 22 '18 at 6:31









          Mostafa Ayaz

          14k3936




          14k3936























              0














              The differentiation & setting of the derivative to zero yields stationary points. A range-of-x ought to have been specified for this question, as without it it is ill-posed.



              To say that the function has a maximum value of ∞ at x=∞ is a scrambling of the meaning of what's going-on, which is that the function increases without limit as x increases without limit, & therefore does not have a maximum value. If a range of x had been given, you would compare the -7 gotten as a local maximum to the value taken by the function at the upper end of the range of x (you need not consider the lower end of the range of x as the function is decreasing without limit in that direction) ... and whichever were the greater would be the answer to the question.



              A fussy little point though: but it's important in many kinds of problem: the range would have to be a closed range - ie of the form $$aleq xleq b ,$$ often denoted $$xin[a,b], $$ rather than what is called an open range $$a< x< b ,$$ often denoted $$xin(a,b) ;$$ as if it were the latter that were specified, you would technically still have the problem of the non-existence of a maximum value of the function if the upper limit of the range were such that it could exceed -7 for x still less than it.






              share|cite|improve this answer




























                0














                The differentiation & setting of the derivative to zero yields stationary points. A range-of-x ought to have been specified for this question, as without it it is ill-posed.



                To say that the function has a maximum value of ∞ at x=∞ is a scrambling of the meaning of what's going-on, which is that the function increases without limit as x increases without limit, & therefore does not have a maximum value. If a range of x had been given, you would compare the -7 gotten as a local maximum to the value taken by the function at the upper end of the range of x (you need not consider the lower end of the range of x as the function is decreasing without limit in that direction) ... and whichever were the greater would be the answer to the question.



                A fussy little point though: but it's important in many kinds of problem: the range would have to be a closed range - ie of the form $$aleq xleq b ,$$ often denoted $$xin[a,b], $$ rather than what is called an open range $$a< x< b ,$$ often denoted $$xin(a,b) ;$$ as if it were the latter that were specified, you would technically still have the problem of the non-existence of a maximum value of the function if the upper limit of the range were such that it could exceed -7 for x still less than it.






                share|cite|improve this answer


























                  0












                  0








                  0






                  The differentiation & setting of the derivative to zero yields stationary points. A range-of-x ought to have been specified for this question, as without it it is ill-posed.



                  To say that the function has a maximum value of ∞ at x=∞ is a scrambling of the meaning of what's going-on, which is that the function increases without limit as x increases without limit, & therefore does not have a maximum value. If a range of x had been given, you would compare the -7 gotten as a local maximum to the value taken by the function at the upper end of the range of x (you need not consider the lower end of the range of x as the function is decreasing without limit in that direction) ... and whichever were the greater would be the answer to the question.



                  A fussy little point though: but it's important in many kinds of problem: the range would have to be a closed range - ie of the form $$aleq xleq b ,$$ often denoted $$xin[a,b], $$ rather than what is called an open range $$a< x< b ,$$ often denoted $$xin(a,b) ;$$ as if it were the latter that were specified, you would technically still have the problem of the non-existence of a maximum value of the function if the upper limit of the range were such that it could exceed -7 for x still less than it.






                  share|cite|improve this answer














                  The differentiation & setting of the derivative to zero yields stationary points. A range-of-x ought to have been specified for this question, as without it it is ill-posed.



                  To say that the function has a maximum value of ∞ at x=∞ is a scrambling of the meaning of what's going-on, which is that the function increases without limit as x increases without limit, & therefore does not have a maximum value. If a range of x had been given, you would compare the -7 gotten as a local maximum to the value taken by the function at the upper end of the range of x (you need not consider the lower end of the range of x as the function is decreasing without limit in that direction) ... and whichever were the greater would be the answer to the question.



                  A fussy little point though: but it's important in many kinds of problem: the range would have to be a closed range - ie of the form $$aleq xleq b ,$$ often denoted $$xin[a,b], $$ rather than what is called an open range $$a< x< b ,$$ often denoted $$xin(a,b) ;$$ as if it were the latter that were specified, you would technically still have the problem of the non-existence of a maximum value of the function if the upper limit of the range were such that it could exceed -7 for x still less than it.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 22 '18 at 8:04

























                  answered Nov 22 '18 at 5:39









                  AmbretteOrrisey

                  57410




                  57410






























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