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- f(x)= -5cos(cos(x5)) f'(x)=?1. Given y = 4 cos(x) - x. Determine where the function is increasing in [-2, 7].Find all values of x in the interval [ -pi/2, pi/2 ]at which the fuction f(x)=sin^2+cos x reaches an absolute minimum value. Note: The only x values in this interval where f'(x)=0 are x= - pi/3 and x= pi/3
- Show that cos x = x has a solution in the interval [0, 1]. Hint: Show that f (x) = x − cos x has a zero in [0, 1].f(x) is a sin function with amplitute 1, and is such that f(0)=0 f(1)=0 f(x) >= 0 for x element [0,1] a) f(x) = ?Let f (x) = x sin x and g(x) = x cos x. (a) Show that f ,(x) = g(x) + sin x and g ,(x) = −f (x) + cos x. (b) Verify that f ,,(x) = −f (x) + 2 cos x and g ,,(x) = −g(x) − 2 sin x. (c) By further experimentation, try to find formulas for all higher derivatives of f and g. Hint: The kth derivative depends on whether k = 4n, 4n + 1, 4n + 2, or 4n + 3.