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f(x) = (2x + 5)/(x² – 4) on [-5,–3].](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff37f5509-02a5-4162-bac3-bb47858168cc%2F4779d50b-6b74-4990-93ef-199bbed755c4%2Fu19rh4h_processed.jpeg&w=3840&q=75)
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- Find the general solution of f′(x) = −2x sin x2 .Find all the x-intercepts of f(x)=cos(x) on the interval [0,2π)(a) From sin2 x + cos2 x = 1, we have f(x) + g(x) = 1. Take the derivative of both sides of this equation to obtain f' (x) + g' (x) = 0. This implies f'(x) = -g'(x). (b) f' (x) = 2 sin x cos x, and g'(x) = 2(cosx)(-sinx) = -2sin x cos x. So, f'(x) = -g'(x).
- If f(x)= (x^4) + 3x + 1 and there are two roots x=a and x=b in the interval [-3,-1] with a <b. Then f(a) =f(b) = ________ ? Since f is continuous on the interval [-3,-1] and differentiable on the interval (-3,-1), by Rolle's Theorem there would exist a point c in interval (a, b) so that f'(c)= 0. However, the only solution to f '(x)= 0 is x= ________ , which is not in the interval (a,b) or in [-3,-1]. Thus, f cannot have more than one root in [-3,-1]. What would the blank be?Solve for the 4th derivative of function K K= x3Cos√xLet 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.