Finding absolute and relative extrema of a function
I don't know if what I'm doing is correct.
Let $f:mathbb{R} to mathbb{R}$ be a function, $f(x)=e^{x^4-3x^2}$.
Choose the correct answer:
$f$ has an absolute maximum at $x=0$.
$f$ has a relative maximum at $x=0$ which isn't absolute.
$f$ has a absolute minimum at $x=1$ and doesn't have any relative minimum. (Does this even make sense?)
$f$ doesn't have relative maximum.
What I've been doing:
Since the question doesn't tell me the interval where I have to find the relative extrema, by the options the gave me I suppose that's $I=[0,1]$.
I know that when I have to find the relative extrema, I have to find the critical points for $f$ at the given interval, but this function doesn't have any. So how do I find it? (Algebraiclly).
When I look at the function it clearly does have a relative maximum at $x=0$ (Option 2), and it's clearly not absolute as $lim_{x to +infty} f(x) = +infty$ (is finding the limit as $x to +infty$ correct when trying to find the absolute extrema?).
So, how do I prove that option 2 is correct without looking at the graphic?
calculus real-analysis
asked Nov 22 '18 at 16:46
parishiltonparishilton
19310
add a comment |
I don't know if what I'm doing is correct.
Let $f:mathbb{R} to mathbb{R}$ be a function, $f(x)=e^{x^4-3x^2}$.
Choose the correct answer:
$f$ has an absolute maximum at $x=0$.
$f$ has a relative maximum at $x=0$ which isn't absolute.
$f$ has a absolute minimum at $x=1$ and doesn't have any relative minimum. (Does this even make sense?)
$f$ doesn't have relative maximum.
What I've been doing:
Since the question doesn't tell me the interval where I have to find the relative extrema, by the options the gave me I suppose that's $I=[0,1]$.
I know that when I have to find the relative extrema, I have to find the critical points for $f$ at the given interval, but this function doesn't have any. So how do I find it? (Algebraiclly).
When I look at the function it clearly does have a relative maximum at $x=0$ (Option 2), and it's clearly not absolute as $lim_{x to +infty} f(x) = +infty$ (is finding the limit as $x to +infty$ correct when trying to find the absolute extrema?).
So, how do I prove that option 2 is correct without looking at the graphic?
calculus real-analysis
asked Nov 22 '18 at 16:46
parishiltonparishilton
19310
I don't understand you approach. First, you tell us that the domain of $f$ is $mathbb R$. Then you write that “the question doesn't tell me the interval where I have to find the relative extrema”. Finally, you decree that that interval is $[0,1]$.
– José Carlos Santos
Nov 22 '18 at 16:49
Yes it's kinda confussing, I thought that the interval was $[0,1]$ by the options they gave me. So I thought that I had to find the relative extrema at that interval.
– parishilton
Nov 22 '18 at 16:52
add a comment |
I don't know if what I'm doing is correct.
Let $f:mathbb{R} to mathbb{R}$ be a function, $f(x)=e^{x^4-3x^2}$.
Choose the correct answer:
$f$ has an absolute maximum at $x=0$.
$f$ has a relative maximum at $x=0$ which isn't absolute.
$f$ has a absolute minimum at $x=1$ and doesn't have any relative minimum. (Does this even make sense?)
$f$ doesn't have relative maximum.
What I've been doing:
Since the question doesn't tell me the interval where I have to find the relative extrema, by the options the gave me I suppose that's $I=[0,1]$.
I know that when I have to find the relative extrema, I have to find the critical points for $f$ at the given interval, but this function doesn't have any. So how do I find it? (Algebraiclly).
When I look at the function it clearly does have a relative maximum at $x=0$ (Option 2), and it's clearly not absolute as $lim_{x to +infty} f(x) = +infty$ (is finding the limit as $x to +infty$ correct when trying to find the absolute extrema?).
So, how do I prove that option 2 is correct without looking at the graphic?
calculus real-analysis
asked Nov 22 '18 at 16:46
parishiltonparishilton
19310
I don't know if what I'm doing is correct.
Let $f:mathbb{R} to mathbb{R}$ be a function, $f(x)=e^{x^4-3x^2}$.
Choose the correct answer:
$f$ has an absolute maximum at $x=0$.
$f$ has a relative maximum at $x=0$ which isn't absolute.
$f$ has a absolute minimum at $x=1$ and doesn't have any relative minimum. (Does this even make sense?)
$f$ doesn't have relative maximum.
What I've been doing:
Since the question doesn't tell me the interval where I have to find the relative extrema, by the options the gave me I suppose that's $I=[0,1]$.
I know that when I have to find the relative extrema, I have to find the critical points for $f$ at the given interval, but this function doesn't have any. So how do I find it? (Algebraiclly).
When I look at the function it clearly does have a relative maximum at $x=0$ (Option 2), and it's clearly not absolute as $lim_{x to +infty} f(x) = +infty$ (is finding the limit as $x to +infty$ correct when trying to find the absolute extrema?).
So, how do I prove that option 2 is correct without looking at the graphic?
calculus real-analysis
calculus real-analysis
asked Nov 22 '18 at 16:46
parishiltonparishilton
19310
asked Nov 22 '18 at 16:46
parishiltonparishilton
19310
asked Nov 22 '18 at 16:46
parishiltonparishilton
19310
asked Nov 22 '18 at 16:46
parishiltonparishilton
19310
asked Nov 22 '18 at 16:46
parishiltonparishilton
19310
19310
I don't understand you approach. First, you tell us that the domain of $f$ is $mathbb R$. Then you write that “the question doesn't tell me the interval where I have to find the relative extrema”. Finally, you decree that that interval is $[0,1]$.
– José Carlos Santos
Nov 22 '18 at 16:49
Yes it's kinda confussing, I thought that the interval was $[0,1]$ by the options they gave me. So I thought that I had to find the relative extrema at that interval.
– parishilton
Nov 22 '18 at 16:52
add a comment |
I don't understand you approach. First, you tell us that the domain of $f$ is $mathbb R$. Then you write that “the question doesn't tell me the interval where I have to find the relative extrema”. Finally, you decree that that interval is $[0,1]$.
– José Carlos Santos
Nov 22 '18 at 16:49
Yes it's kinda confussing, I thought that the interval was $[0,1]$ by the options they gave me. So I thought that I had to find the relative extrema at that interval.
– parishilton
Nov 22 '18 at 16:52
I don't understand you approach. First, you tell us that the domain of $f$ is $mathbb R$. Then you write that “the question doesn't tell me the interval where I have to find the relative extrema”. Finally, you decree that that interval is $[0,1]$.
– José Carlos Santos
Nov 22 '18 at 16:49
I don't understand you approach. First, you tell us that the domain of $f$ is $mathbb R$. Then you write that “the question doesn't tell me the interval where I have to find the relative extrema”. Finally, you decree that that interval is $[0,1]$.
– José Carlos Santos
Nov 22 '18 at 16:49
Yes it's kinda confussing, I thought that the interval was $[0,1]$ by the options they gave me. So I thought that I had to find the relative extrema at that interval.
– parishilton
Nov 22 '18 at 16:52
Yes it's kinda confussing, I thought that the interval was $[0,1]$ by the options they gave me. So I thought that I had to find the relative extrema at that interval.
– parishilton
Nov 22 '18 at 16:52
add a comment |
1 Answer
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HINT
We have that
$$f(x)=e^{x^4-3x^2}>0 implies f'(x)=(4x^3-6x)e^{x^4-3x^2}=0 implies x=0 ,lor , x=pm sqrt{frac23}$$
then consider the sign of $f'(x)$.
answered Nov 22 '18 at 16:49
gimusigimusi
1
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
HINT
We have that
$$f(x)=e^{x^4-3x^2}>0 implies f'(x)=(4x^3-6x)e^{x^4-3x^2}=0 implies x=0 ,lor , x=pm sqrt{frac23}$$
then consider the sign of $f'(x)$.
answered Nov 22 '18 at 16:49
gimusigimusi
1
add a comment |
HINT
We have that
$$f(x)=e^{x^4-3x^2}>0 implies f'(x)=(4x^3-6x)e^{x^4-3x^2}=0 implies x=0 ,lor , x=pm sqrt{frac23}$$
then consider the sign of $f'(x)$.
answered Nov 22 '18 at 16:49
gimusigimusi
1
add a comment |
HINT
We have that
$$f(x)=e^{x^4-3x^2}>0 implies f'(x)=(4x^3-6x)e^{x^4-3x^2}=0 implies x=0 ,lor , x=pm sqrt{frac23}$$
then consider the sign of $f'(x)$.
answered Nov 22 '18 at 16:49
gimusigimusi
1
HINT
We have that
$$f(x)=e^{x^4-3x^2}>0 implies f'(x)=(4x^3-6x)e^{x^4-3x^2}=0 implies x=0 ,lor , x=pm sqrt{frac23}$$
then consider the sign of $f'(x)$.
answered Nov 22 '18 at 16:49
gimusigimusi
1
answered Nov 22 '18 at 16:49
gimusigimusi
1
answered Nov 22 '18 at 16:49
gimusigimusi
1
answered Nov 22 '18 at 16:49
gimusigimusi
1
1
add a comment |
add a comment |
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I don't understand you approach. First, you tell us that the domain of $f$ is $mathbb R$. Then you write that “the question doesn't tell me the interval where I have to find the relative extrema”. Finally, you decree that that interval is $[0,1]$.
– José Carlos Santos
Nov 22 '18 at 16:49
Yes it's kinda confussing, I thought that the interval was $[0,1]$ by the options they gave me. So I thought that I had to find the relative extrema at that interval.
– parishilton
Nov 22 '18 at 16:52