how many solutions $f(x)=0$ has over an interval given derivative graph
0
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The equation $f(x)=0$ can have no more than two solutions on the interval $[-3,3]$. Explain why. I was given the graph of the derivative with this question, and it looked like $arctan x$ over $(-3,3)$.
calculus
edited Nov 28 '18 at 2:02
Tianlalu
3,08621038
asked Nov 28 '18 at 0:38
nekoBlackSunnekoBlackSun
1
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show 2 more comments
0
$begingroup$
The equation $f(x)=0$ can have no more than two solutions on the interval $[-3,3]$. Explain why. I was given the graph of the derivative with this question, and it looked like $arctan x$ over $(-3,3)$.
calculus
edited Nov 28 '18 at 2:02
Tianlalu
3,08621038
asked Nov 28 '18 at 0:38
nekoBlackSunnekoBlackSun
1
$endgroup$
1
$begingroup$
Can you draw a function with say, three zeros? What can you say about the derivative? Did you perhaps just learn something called the mean value theorem?
$endgroup$
– qbert
Nov 28 '18 at 0:43
$begingroup$
Well, my teacher just gave it to us and didn't teach us the theorem. I searched it up but I am not sure how to apply it.
$endgroup$
– nekoBlackSun
Nov 28 '18 at 0:49
1
$begingroup$
Echoing what @qbert mentioned, suppose (by way of contradiction) that $f(x)$ had $3$ or more distinct zeros on $[-3,4]$. What does the mean value theorem, or even Rolle's Theorem, tell you about the number of zeros that it's derivative has? How many zeros does its derivative actually have?
$endgroup$
– AlkaKadri
Nov 28 '18 at 0:53
$begingroup$
The derivative has one zero judging by the graph. However, I don't have the equation, so I don't think I can use MVT.
$endgroup$
– nekoBlackSun
Nov 28 '18 at 1:10
$begingroup$
I can't see the graph, but how many times do you see the derivative (i.e. the slope of the function) changing sign? Is it just once? If the derivative goes from positive to negative (or vice-versa) only once, you cannot have the function cross the x-axis more than twice (i.e. no more than two solutions). Of course, it is possible to have fewer than two solutions.
$endgroup$
– Aditya Dua
Nov 28 '18 at 1:50
|
show 2 more comments
0
0
0
$begingroup$
The equation $f(x)=0$ can have no more than two solutions on the interval $[-3,3]$. Explain why. I was given the graph of the derivative with this question, and it looked like $arctan x$ over $(-3,3)$.
calculus
edited Nov 28 '18 at 2:02
Tianlalu
3,08621038
asked Nov 28 '18 at 0:38
nekoBlackSunnekoBlackSun
1
$endgroup$
The equation $f(x)=0$ can have no more than two solutions on the interval $[-3,3]$. Explain why. I was given the graph of the derivative with this question, and it looked like $arctan x$ over $(-3,3)$.
calculus
calculus
edited Nov 28 '18 at 2:02
Tianlalu
3,08621038
asked Nov 28 '18 at 0:38
nekoBlackSunnekoBlackSun
1
edited Nov 28 '18 at 2:02
Tianlalu
3,08621038
asked Nov 28 '18 at 0:38
nekoBlackSunnekoBlackSun
1
edited Nov 28 '18 at 2:02
Tianlalu
3,08621038
edited Nov 28 '18 at 2:02
Tianlalu
3,08621038
edited Nov 28 '18 at 2:02
Tianlalu
3,08621038
3,08621038
asked Nov 28 '18 at 0:38
nekoBlackSunnekoBlackSun
1
asked Nov 28 '18 at 0:38
nekoBlackSunnekoBlackSun
1
asked Nov 28 '18 at 0:38
nekoBlackSunnekoBlackSun
1
1
1
$begingroup$
Can you draw a function with say, three zeros? What can you say about the derivative? Did you perhaps just learn something called the mean value theorem?
$endgroup$
– qbert
Nov 28 '18 at 0:43
$begingroup$
Well, my teacher just gave it to us and didn't teach us the theorem. I searched it up but I am not sure how to apply it.
$endgroup$
– nekoBlackSun
Nov 28 '18 at 0:49
1
$begingroup$
Echoing what @qbert mentioned, suppose (by way of contradiction) that $f(x)$ had $3$ or more distinct zeros on $[-3,4]$. What does the mean value theorem, or even Rolle's Theorem, tell you about the number of zeros that it's derivative has? How many zeros does its derivative actually have?
$endgroup$
– AlkaKadri
Nov 28 '18 at 0:53
$begingroup$
The derivative has one zero judging by the graph. However, I don't have the equation, so I don't think I can use MVT.
$endgroup$
– nekoBlackSun
Nov 28 '18 at 1:10
$begingroup$
I can't see the graph, but how many times do you see the derivative (i.e. the slope of the function) changing sign? Is it just once? If the derivative goes from positive to negative (or vice-versa) only once, you cannot have the function cross the x-axis more than twice (i.e. no more than two solutions). Of course, it is possible to have fewer than two solutions.
$endgroup$
– Aditya Dua
Nov 28 '18 at 1:50
|
show 2 more comments
1
$begingroup$
Can you draw a function with say, three zeros? What can you say about the derivative? Did you perhaps just learn something called the mean value theorem?
$endgroup$
– qbert
Nov 28 '18 at 0:43
$begingroup$
Well, my teacher just gave it to us and didn't teach us the theorem. I searched it up but I am not sure how to apply it.
$endgroup$
– nekoBlackSun
Nov 28 '18 at 0:49
1
$begingroup$
Echoing what @qbert mentioned, suppose (by way of contradiction) that $f(x)$ had $3$ or more distinct zeros on $[-3,4]$. What does the mean value theorem, or even Rolle's Theorem, tell you about the number of zeros that it's derivative has? How many zeros does its derivative actually have?
$endgroup$
– AlkaKadri
Nov 28 '18 at 0:53
$begingroup$
The derivative has one zero judging by the graph. However, I don't have the equation, so I don't think I can use MVT.
$endgroup$
– nekoBlackSun
Nov 28 '18 at 1:10
$begingroup$
I can't see the graph, but how many times do you see the derivative (i.e. the slope of the function) changing sign? Is it just once? If the derivative goes from positive to negative (or vice-versa) only once, you cannot have the function cross the x-axis more than twice (i.e. no more than two solutions). Of course, it is possible to have fewer than two solutions.
$endgroup$
– Aditya Dua
Nov 28 '18 at 1:50
1
1
$begingroup$
Can you draw a function with say, three zeros? What can you say about the derivative? Did you perhaps just learn something called the mean value theorem?
$endgroup$
– qbert
Nov 28 '18 at 0:43
$begingroup$
Can you draw a function with say, three zeros? What can you say about the derivative? Did you perhaps just learn something called the mean value theorem?
$endgroup$
– qbert
Nov 28 '18 at 0:43
$begingroup$
Well, my teacher just gave it to us and didn't teach us the theorem. I searched it up but I am not sure how to apply it.
$endgroup$
– nekoBlackSun
Nov 28 '18 at 0:49
$begingroup$
Well, my teacher just gave it to us and didn't teach us the theorem. I searched it up but I am not sure how to apply it.
$endgroup$
– nekoBlackSun
Nov 28 '18 at 0:49
1
1
$begingroup$
Echoing what @qbert mentioned, suppose (by way of contradiction) that $f(x)$ had $3$ or more distinct zeros on $[-3,4]$. What does the mean value theorem, or even Rolle's Theorem, tell you about the number of zeros that it's derivative has? How many zeros does its derivative actually have?
$endgroup$
– AlkaKadri
Nov 28 '18 at 0:53
$begingroup$
Echoing what @qbert mentioned, suppose (by way of contradiction) that $f(x)$ had $3$ or more distinct zeros on $[-3,4]$. What does the mean value theorem, or even Rolle's Theorem, tell you about the number of zeros that it's derivative has? How many zeros does its derivative actually have?
$endgroup$
– AlkaKadri
Nov 28 '18 at 0:53
$begingroup$
The derivative has one zero judging by the graph. However, I don't have the equation, so I don't think I can use MVT.
$endgroup$
– nekoBlackSun
Nov 28 '18 at 1:10
$begingroup$
The derivative has one zero judging by the graph. However, I don't have the equation, so I don't think I can use MVT.
$endgroup$
– nekoBlackSun
Nov 28 '18 at 1:10
$begingroup$
I can't see the graph, but how many times do you see the derivative (i.e. the slope of the function) changing sign? Is it just once? If the derivative goes from positive to negative (or vice-versa) only once, you cannot have the function cross the x-axis more than twice (i.e. no more than two solutions). Of course, it is possible to have fewer than two solutions.
$endgroup$
– Aditya Dua
Nov 28 '18 at 1:50
$begingroup$
I can't see the graph, but how many times do you see the derivative (i.e. the slope of the function) changing sign? Is it just once? If the derivative goes from positive to negative (or vice-versa) only once, you cannot have the function cross the x-axis more than twice (i.e. no more than two solutions). Of course, it is possible to have fewer than two solutions.
$endgroup$
– Aditya Dua
Nov 28 '18 at 1:50
|
show 2 more comments
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1
$begingroup$
Can you draw a function with say, three zeros? What can you say about the derivative? Did you perhaps just learn something called the mean value theorem?
$endgroup$
– qbert
Nov 28 '18 at 0:43
$begingroup$
Well, my teacher just gave it to us and didn't teach us the theorem. I searched it up but I am not sure how to apply it.
$endgroup$
– nekoBlackSun
Nov 28 '18 at 0:49
1
$begingroup$
Echoing what @qbert mentioned, suppose (by way of contradiction) that $f(x)$ had $3$ or more distinct zeros on $[-3,4]$. What does the mean value theorem, or even Rolle's Theorem, tell you about the number of zeros that it's derivative has? How many zeros does its derivative actually have?
$endgroup$
– AlkaKadri
Nov 28 '18 at 0:53
$begingroup$
The derivative has one zero judging by the graph. However, I don't have the equation, so I don't think I can use MVT.
$endgroup$
– nekoBlackSun
Nov 28 '18 at 1:10
$begingroup$
I can't see the graph, but how many times do you see the derivative (i.e. the slope of the function) changing sign? Is it just once? If the derivative goes from positive to negative (or vice-versa) only once, you cannot have the function cross the x-axis more than twice (i.e. no more than two solutions). Of course, it is possible to have fewer than two solutions.
$endgroup$
– Aditya Dua
Nov 28 '18 at 1:50