Question

Find the linearization of each function at the indicated value using the derivative provided. (a) f(x)=sinx at x= 2π​ if f ′(x)=cos(x) (b) f(x)=x 3/2 at x=16 if f ′ (x)= 23 x 1/2 (c) f(x)=2 x at x=0 if f ′ (x)=ln(2)2 x

Answer 1

(a) The linearization of f(x) = sin(x) at x = 2π is L(x) = √3/2 + (x - 2π).

(b) The linearization of f(x) = x^(3/2) at x = 16 is L(x) = 96 + 3√2(x - 16).

(c) The linearization of f(x) = 2x at x = 0 is L(x) = 0 + ln(2)x.

Step-by-step explanation:

(a) To find the linearization of f(x) = sin(x) at x = 2π, we use the formula for linearization: L(x) = f(a) + f'(a)(x - a). Here, a = 2π. The derivative of f(x) = sin(x) is f'(x) = cos(x). At x = 2π, f'(2π) = cos(2π) = 1. Therefore, L(x) = sin(2π) + cos(2π)(x - 2π) = 0 + 1(x - 2π) = x - 2π.

(b) For f(x) = x^(3/2) at x = 16, the derivative f'(x) = 23x^(1/2). At x = 16, f'(16) = 23 * 16^(1/2) = 23 * 4 = 92. Using the linearization formula, L(x) = f(16) + f'(16)(x - 16) = 16^(3/2) + 92(x - 16) = 96 + 3√2(x - 16).

(c) Lastly, for f(x) = 2x at x = 0, the derivative f'(x) = ln(2) * 2x. At x = 0, f'(0) = ln(2) * 2 * 0 = 0. Applying the linearization formula, L(x) = f(0) + f'(0)(x - 0) = 2 * 0 + 0(x - 0) = 0 + ln(2)x = ln(2)x.

These linearizations are approximations of the original functions around the given points, useful for estimating values of the functions close to those points using simpler linear equations.