Find the derivative of sin(2cosx)
Find the derivative of sin(2cosx)

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.

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

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

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

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