Final answer:
The linearization of the function g(x) = xf(x²) at x=2, given the provided function values and derivatives, is l(x) = 10 - 3(x - 2).
Step-by-step explanation:
To find the linearization l(x) of the function g(x) = xf(x²) at x=2, we will use the concept of the derivative to find the slope of the tangent line at this point, and hence determine the linear approximation.
First, let's determine g'(x) using the product rule and the chain rule:
- Differentiate g(x) with respect to x: g'(x) = f(x²) + 2x(f'(x²)).
- Evaluate g'(x) at x=2 using the given values: g'(2) = f(4) + 2(2)f'(4) = 5 + 4(-2) = 5 - 8 = -3.
- The linearization is of the form l(x) = g(a) + g'(a)(x - a), where a=2.
- Since g(2) = 2f(4) = 2(5) = 10 and g'(2) = -3, we get l(x) = 10 - 3(x - 2).
The linear approximation or linearization of g(x) at x=2 is therefore l(x) = 10 - 3(x - 2).