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The second derivative of the function f is given by f′′(x)=sin((x^2)/8)−2cosx. The function f has many critical points, two of which are at x=0 and x=6.949. Which of the following statements is true?

 

a) f has a local minimum at x=0 and at x=6.949.

b) f has a local minimum at x=0 and a local maximum at x=6.949.

c) f has a local maximum at x=0 and a local minimum at x=6.949.

d) f has a local maximum at x=0 and at x=6.949.

 Dec 15, 2020

Best Answer 

 #1
avatar+9485 
+1

f(x) = sin(x28)2cos(x)

 

Let's plug in each of the given critical x-values into the second derivative to see whether it is positive or negative.

 

f(0) = sin(028)2cos(0) = sin(0)2cos(0) = 02(1) = 2 < 0

 

Since  f''(0) < 0  the graph of  f  is concave down at  x = 0  and so a local max occurs at  x = 0

 

f(6.949) = sin(6.94928)2cos(6.949)  1.817 < 0    (assuming  x  is in radians)

 

Since  f''(6.949) < 0  the graph of  f  is concave down at  x = 6.949  and so a local max occurs at  x = 6.949

 

Here's some more info about this:

 

https://www.khanacademy.org/math/ap-calculus-ab/ab-diff-analytical-applications-new/ab-5-7/v/second-derivative-test

 Dec 16, 2020
 #1
avatar+9485 
+1
Best Answer

f(x) = sin(x28)2cos(x)

 

Let's plug in each of the given critical x-values into the second derivative to see whether it is positive or negative.

 

f(0) = sin(028)2cos(0) = sin(0)2cos(0) = 02(1) = 2 < 0

 

Since  f''(0) < 0  the graph of  f  is concave down at  x = 0  and so a local max occurs at  x = 0

 

f(6.949) = sin(6.94928)2cos(6.949)  1.817 < 0    (assuming  x  is in radians)

 

Since  f''(6.949) < 0  the graph of  f  is concave down at  x = 6.949  and so a local max occurs at  x = 6.949

 

Here's some more info about this:

 

https://www.khanacademy.org/math/ap-calculus-ab/ab-diff-analytical-applications-new/ab-5-7/v/second-derivative-test

hectictar Dec 16, 2020

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