Suppose f and g are functions that are differentiable at x = 1 and that f(1) = 2, f '(1) = -1, g(1) = -2, and g '(1) = 3. Find the value of h '(1).

h(x) = (x2 + 11) g(x)
h '(1) =

To find the derivative of h(x), we can use the product rule. The product rule states that if we have two functions, f(x) and g(x), then the derivative of their product, h(x) = f(x) * g(x), is given by:

h'(x) = f'(x) * g(x) + f(x) * g'(x)

Given that h(x) = (x^2 + 11) * g(x), we can now apply the product rule to find h'(x).

First, let's find f(x) and f'(x). We are given that f(1) = 2 and f'(1) = -1, so f(x) must be equal to x + b for some constant b, and f'(x) must be equal to 1.

Using the fact that f(1) = 2, we can substitute x = 1 into f(x) to find b:

f(1) = 1 + b = 2
b = 2 - 1
b = 1

So, f(x) = x + 1 and f'(x) = 1.

Now, let's substitute f(x), f'(x), g(x), and g'(x) into the product rule formula for h'(x):

h'(x) = (x^2 + 11) * g'(x) + (x + 1) * g(x)

Substituting x = 1, we get:

h'(1) = (1^2 + 11) * g'(1) + (1 + 1) * g(1)
= 12 * 3 + 2 * (-2)
= 36 - 4
= 32

Therefore, h'(1) = 32.