The table about give values of a twice differentiable function f and its first derivative f' for selected values of x. Let g be the function defined by g(x)= f (x2-x).

Question

Unless otherwise specified, the domain of a function f is assumed to be the set of all real numbers x for which f(X) is a real number

x. -3 0 3 6

f(x) -5 4 1 7

f'(x) -1 2 -2 4

The table about give values of a twice differentiable function f and its first derivative f' for selected values of x. Let g be the function defined by g(x)= f (x²-x).

a) What is the value of g'(3)?

b) It is know that g''(0)= -1. What is the value of f ''(0)?

c) Is there a value c, for 0<c<3, such that g(c)=5?

Answer 1

Solution:

(a) By the chain rule,

g'(x) = f'(x^2 - x)*(2x - 1)

So g'(3) = f'(3^2 - 3)*(2*3 - 1)

= 5 * f'(6)

= 5 * 4

= 20 ← Answer

(b) Now by the product rule and the chain rule,

g''(x) = f''(x^2 - x)*(2x-1)^2 + f'(x^2 - x)*(2)

g''(0) = f''(0)*(-1)^2 + f'(0)*(2)

From the given g''(0) = -1 we have

-1 = f''(0) + 2*(2)

f''(0) = -5 ← Answer

(c) From the table,

g(0) = f(0^2 - 0) = f(0) = 4

g(3) = f(3^2 - 3) = f(6) = 7

Since f is continuous on R, g(x) = f(x^2 -x) is continuous on R.

Now, 4 = g(0) < 5 < g(3) = 7, so by the Intermediate Value Theorem,

there exists c with 0 < c < 3 such that g(c) = 5.