+234 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.
+9481 \(f''(x)\ =\ \sin(\frac{x^2}{8})-2\cos(x) \)
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
Here's some more info about this:
+9481 \(f''(x)\ =\ \sin(\frac{x^2}{8})-2\cos(x) \)
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
Here's some more info about this:
