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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

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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

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$$f''(x)\ =\ \sin(\frac{x^2}{8})-2\cos(x)$$

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

$$f''(0)\ =\ \sin(\frac{0^2}{8})-2\cos(0)\ =\ \sin(0)-2\cos(0)\ =\ 0-2(1)\ =\ -2\ <\ 0$$

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

$$f''(6.949)\ =\ \sin(\frac{6.949^2}{8})-2\cos(6.949)\ \approx\ -1.817 \ <\ 0$$    (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

Dec 16, 2020

#1
+9277
+1

$$f''(x)\ =\ \sin(\frac{x^2}{8})-2\cos(x)$$

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

$$f''(0)\ =\ \sin(\frac{0^2}{8})-2\cos(0)\ =\ \sin(0)-2\cos(0)\ =\ 0-2(1)\ =\ -2\ <\ 0$$

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

$$f''(6.949)\ =\ \sin(\frac{6.949^2}{8})-2\cos(6.949)\ \approx\ -1.817 \ <\ 0$$    (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