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Calculus: Determining function from the graph of its derivative [closed]
The figure(attached below) shows the graph of df/dx , the derivative of a twice differentiable function f , on the closed interval 0<= x <= 8. The graph of df/dx has horizontal tangent lines at x=1,x=3 , and x=5 . The function f is defined for all real numbers.
Answer the following questions.
1)On which open interval(s) contained in 0< x < 8 is the graph of f both concave down and decreasing (simultaneously)?(integer value)
2)The function g is defined by g(x) = (f(x))^3 . If f(3) = -5/2, find the slope of the line tangent to the graph of g at x=3 .
I don't know how to precede with them, please help.
calculus
edited Nov 20 '18 at 16:35
Xander Henderson
14.1k103554
asked Nov 20 '18 at 16:31
J.Doe
1
closed as off-topic by Xander Henderson, John Douma, darij grinberg, Chinnapparaj R, John B Nov 21 '18 at 10:32
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Xander Henderson, John Douma, darij grinberg, Chinnapparaj R, John B
If this question can be reworded to fit the rules in the help center, please edit the question.
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The figure(attached below) shows the graph of df/dx , the derivative of a twice differentiable function f , on the closed interval 0<= x <= 8. The graph of df/dx has horizontal tangent lines at x=1,x=3 , and x=5 . The function f is defined for all real numbers.
Answer the following questions.
1)On which open interval(s) contained in 0< x < 8 is the graph of f both concave down and decreasing (simultaneously)?(integer value)
2)The function g is defined by g(x) = (f(x))^3 . If f(3) = -5/2, find the slope of the line tangent to the graph of g at x=3 .
I don't know how to precede with them, please help.
calculus
edited Nov 20 '18 at 16:35
Xander Henderson
14.1k103554
asked Nov 20 '18 at 16:31
J.Doe
1
closed as off-topic by Xander Henderson, John Douma, darij grinberg, Chinnapparaj R, John B Nov 21 '18 at 10:32
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Xander Henderson, John Douma, darij grinberg, Chinnapparaj R, John B
If this question can be reworded to fit the rules in the help center, please edit the question.
Hint: if $f$ starts increasing when the blue curve passes through a root, then $f$ flips and is decreasing. So you have find the intervals where $f$ is increasing and decreasing, and try to find where in the intervals the function is concave down.
– ja72
Nov 20 '18 at 17:22
add a comment |
The figure(attached below) shows the graph of df/dx , the derivative of a twice differentiable function f , on the closed interval 0<= x <= 8. The graph of df/dx has horizontal tangent lines at x=1,x=3 , and x=5 . The function f is defined for all real numbers.
Answer the following questions.
1)On which open interval(s) contained in 0< x < 8 is the graph of f both concave down and decreasing (simultaneously)?(integer value)
2)The function g is defined by g(x) = (f(x))^3 . If f(3) = -5/2, find the slope of the line tangent to the graph of g at x=3 .
I don't know how to precede with them, please help.
calculus
edited Nov 20 '18 at 16:35
Xander Henderson
14.1k103554
asked Nov 20 '18 at 16:31
J.Doe
1
The figure(attached below) shows the graph of df/dx , the derivative of a twice differentiable function f , on the closed interval 0<= x <= 8. The graph of df/dx has horizontal tangent lines at x=1,x=3 , and x=5 . The function f is defined for all real numbers.
Answer the following questions.
1)On which open interval(s) contained in 0< x < 8 is the graph of f both concave down and decreasing (simultaneously)?(integer value)
2)The function g is defined by g(x) = (f(x))^3 . If f(3) = -5/2, find the slope of the line tangent to the graph of g at x=3 .
I don't know how to precede with them, please help.
calculus
calculus
edited Nov 20 '18 at 16:35
Xander Henderson
14.1k103554
asked Nov 20 '18 at 16:31
J.Doe
1
edited Nov 20 '18 at 16:35
Xander Henderson
14.1k103554
asked Nov 20 '18 at 16:31
J.Doe
1
edited Nov 20 '18 at 16:35
Xander Henderson
14.1k103554
edited Nov 20 '18 at 16:35
Xander Henderson
14.1k103554
edited Nov 20 '18 at 16:35
Xander Henderson
14.1k103554
14.1k103554
asked Nov 20 '18 at 16:31
J.Doe
1
asked Nov 20 '18 at 16:31
J.Doe
1
asked Nov 20 '18 at 16:31
J.Doe
1
1
closed as off-topic by Xander Henderson, John Douma, darij grinberg, Chinnapparaj R, John B Nov 21 '18 at 10:32
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Xander Henderson, John Douma, darij grinberg, Chinnapparaj R, John B
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Xander Henderson, John Douma, darij grinberg, Chinnapparaj R, John B Nov 21 '18 at 10:32
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Xander Henderson, John Douma, darij grinberg, Chinnapparaj R, John B
If this question can be reworded to fit the rules in the help center, please edit the question.
Hint: if $f$ starts increasing when the blue curve passes through a root, then $f$ flips and is decreasing. So you have find the intervals where $f$ is increasing and decreasing, and try to find where in the intervals the function is concave down.
– ja72
Nov 20 '18 at 17:22
add a comment |
Hint: if $f$ starts increasing when the blue curve passes through a root, then $f$ flips and is decreasing. So you have find the intervals where $f$ is increasing and decreasing, and try to find where in the intervals the function is concave down.
– ja72
Nov 20 '18 at 17:22
Hint: if $f$ starts increasing when the blue curve passes through a root, then $f$ flips and is decreasing. So you have find the intervals where $f$ is increasing and decreasing, and try to find where in the intervals the function is concave down.
– ja72
Nov 20 '18 at 17:22
Hint: if $f$ starts increasing when the blue curve passes through a root, then $f$ flips and is decreasing. So you have find the intervals where $f$ is increasing and decreasing, and try to find where in the intervals the function is concave down.
– ja72
Nov 20 '18 at 17:22
add a comment |
2 Answers
2
active
oldest
votes
$f(x)$ is decreasing if $f'(x) <0$
The function is concave down if $f''(x)<0$ Based on the graph how do you eyeball where $f''(x)<0$?
$g(x) = f^3(x)\
g'(x) = 3f^2(x)f'(x)$
answered Nov 20 '18 at 16:54
Doug M
44k31854
add a comment |
Something of critical importance here is that you are not actually determining the graph of the function $f(x)$; while we are going to say some things about what it is doing, we at no point have to draw anything. In fact, except for the final step in your 2nd question, we won't even need to compute any specific values of $f(x)$ or $f'(x)$ at all.
For question 1, what is the definition of "decreasing" and "concave down"? Since our function is twice differentiable we can use the first and second derivative tests to characterize these two concepts. Decreasing means that the first derivative is negative, and concave down means that the 2nd derivative is negative. The big trick to stitching these two facts together will be that the first derivative $f'(x)$ must be decreasing when the second derivative $f''(x)$ is negative, since the second derivative of $f$ is just the derivative of the first derivative. (What a mouthful!)
For question 2, you are given a new function $g(x)=f(x)^3$ and asked to find the slope of the tangent line at $x=3$. Well, what is the slope of the tangent line? It's the value of the derivative of $g$ at that point. So take the derivative! You will have to use the chain rule to figure this one out, and at the end you will plug in $x=3$ everywhere. We have been told that $f(3)=frac{-5}{2}$, and we can read off the value of $f'(3)$ from the picture you provided.
Hopefully this helps!
answered Nov 20 '18 at 16:48
Valborg
111
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$f(x)$ is decreasing if $f'(x) <0$
The function is concave down if $f''(x)<0$ Based on the graph how do you eyeball where $f''(x)<0$?
$g(x) = f^3(x)\
g'(x) = 3f^2(x)f'(x)$
answered Nov 20 '18 at 16:54
Doug M
44k31854
add a comment |
$f(x)$ is decreasing if $f'(x) <0$
The function is concave down if $f''(x)<0$ Based on the graph how do you eyeball where $f''(x)<0$?
$g(x) = f^3(x)\
g'(x) = 3f^2(x)f'(x)$
answered Nov 20 '18 at 16:54
Doug M
44k31854
add a comment |
$f(x)$ is decreasing if $f'(x) <0$
The function is concave down if $f''(x)<0$ Based on the graph how do you eyeball where $f''(x)<0$?
$g(x) = f^3(x)\
g'(x) = 3f^2(x)f'(x)$
answered Nov 20 '18 at 16:54
Doug M
44k31854
$f(x)$ is decreasing if $f'(x) <0$
The function is concave down if $f''(x)<0$ Based on the graph how do you eyeball where $f''(x)<0$?
$g(x) = f^3(x)\
g'(x) = 3f^2(x)f'(x)$
answered Nov 20 '18 at 16:54
Doug M
44k31854
answered Nov 20 '18 at 16:54
Doug M
44k31854
answered Nov 20 '18 at 16:54
Doug M
44k31854
answered Nov 20 '18 at 16:54
Doug M
44k31854
44k31854
add a comment |
add a comment |
Something of critical importance here is that you are not actually determining the graph of the function $f(x)$; while we are going to say some things about what it is doing, we at no point have to draw anything. In fact, except for the final step in your 2nd question, we won't even need to compute any specific values of $f(x)$ or $f'(x)$ at all.
For question 1, what is the definition of "decreasing" and "concave down"? Since our function is twice differentiable we can use the first and second derivative tests to characterize these two concepts. Decreasing means that the first derivative is negative, and concave down means that the 2nd derivative is negative. The big trick to stitching these two facts together will be that the first derivative $f'(x)$ must be decreasing when the second derivative $f''(x)$ is negative, since the second derivative of $f$ is just the derivative of the first derivative. (What a mouthful!)
For question 2, you are given a new function $g(x)=f(x)^3$ and asked to find the slope of the tangent line at $x=3$. Well, what is the slope of the tangent line? It's the value of the derivative of $g$ at that point. So take the derivative! You will have to use the chain rule to figure this one out, and at the end you will plug in $x=3$ everywhere. We have been told that $f(3)=frac{-5}{2}$, and we can read off the value of $f'(3)$ from the picture you provided.
Hopefully this helps!
answered Nov 20 '18 at 16:48
Valborg
111
add a comment |
Something of critical importance here is that you are not actually determining the graph of the function $f(x)$; while we are going to say some things about what it is doing, we at no point have to draw anything. In fact, except for the final step in your 2nd question, we won't even need to compute any specific values of $f(x)$ or $f'(x)$ at all.
For question 1, what is the definition of "decreasing" and "concave down"? Since our function is twice differentiable we can use the first and second derivative tests to characterize these two concepts. Decreasing means that the first derivative is negative, and concave down means that the 2nd derivative is negative. The big trick to stitching these two facts together will be that the first derivative $f'(x)$ must be decreasing when the second derivative $f''(x)$ is negative, since the second derivative of $f$ is just the derivative of the first derivative. (What a mouthful!)
For question 2, you are given a new function $g(x)=f(x)^3$ and asked to find the slope of the tangent line at $x=3$. Well, what is the slope of the tangent line? It's the value of the derivative of $g$ at that point. So take the derivative! You will have to use the chain rule to figure this one out, and at the end you will plug in $x=3$ everywhere. We have been told that $f(3)=frac{-5}{2}$, and we can read off the value of $f'(3)$ from the picture you provided.
Hopefully this helps!
answered Nov 20 '18 at 16:48
Valborg
111
add a comment |
Something of critical importance here is that you are not actually determining the graph of the function $f(x)$; while we are going to say some things about what it is doing, we at no point have to draw anything. In fact, except for the final step in your 2nd question, we won't even need to compute any specific values of $f(x)$ or $f'(x)$ at all.
For question 1, what is the definition of "decreasing" and "concave down"? Since our function is twice differentiable we can use the first and second derivative tests to characterize these two concepts. Decreasing means that the first derivative is negative, and concave down means that the 2nd derivative is negative. The big trick to stitching these two facts together will be that the first derivative $f'(x)$ must be decreasing when the second derivative $f''(x)$ is negative, since the second derivative of $f$ is just the derivative of the first derivative. (What a mouthful!)
For question 2, you are given a new function $g(x)=f(x)^3$ and asked to find the slope of the tangent line at $x=3$. Well, what is the slope of the tangent line? It's the value of the derivative of $g$ at that point. So take the derivative! You will have to use the chain rule to figure this one out, and at the end you will plug in $x=3$ everywhere. We have been told that $f(3)=frac{-5}{2}$, and we can read off the value of $f'(3)$ from the picture you provided.
Hopefully this helps!
answered Nov 20 '18 at 16:48
Valborg
111
Something of critical importance here is that you are not actually determining the graph of the function $f(x)$; while we are going to say some things about what it is doing, we at no point have to draw anything. In fact, except for the final step in your 2nd question, we won't even need to compute any specific values of $f(x)$ or $f'(x)$ at all.
For question 1, what is the definition of "decreasing" and "concave down"? Since our function is twice differentiable we can use the first and second derivative tests to characterize these two concepts. Decreasing means that the first derivative is negative, and concave down means that the 2nd derivative is negative. The big trick to stitching these two facts together will be that the first derivative $f'(x)$ must be decreasing when the second derivative $f''(x)$ is negative, since the second derivative of $f$ is just the derivative of the first derivative. (What a mouthful!)
For question 2, you are given a new function $g(x)=f(x)^3$ and asked to find the slope of the tangent line at $x=3$. Well, what is the slope of the tangent line? It's the value of the derivative of $g$ at that point. So take the derivative! You will have to use the chain rule to figure this one out, and at the end you will plug in $x=3$ everywhere. We have been told that $f(3)=frac{-5}{2}$, and we can read off the value of $f'(3)$ from the picture you provided.
Hopefully this helps!
answered Nov 20 '18 at 16:48
Valborg
111
answered Nov 20 '18 at 16:48
Valborg
111
answered Nov 20 '18 at 16:48
Valborg
111
answered Nov 20 '18 at 16:48
Valborg
111
111
add a comment |
add a comment |
Hint: if $f$ starts increasing when the blue curve passes through a root, then $f$ flips and is decreasing. So you have find the intervals where $f$ is increasing and decreasing, and try to find where in the intervals the function is concave down.
– ja72
Nov 20 '18 at 17:22