Does the function satisfy mean value theorem?

Question

Determine whether or not each function satisfies the hypothesis for the mean value theorem on the given interval [a,b]. If it does, use derivatives and algebra to fed the exact values of all c € (a,b) that satisfy the conclusion of the mean value theorem.

F(x)= (x^2 -1)(x^2- 4), [a,b] = [-3,3]

Answer 1

Mean value theorem:

If f is

(1) Continuous on closed interval [a,b] where a < b

(2) Differentiable on the open interval (a,b)

then there exist at least one point c in the (a, b) such that .

Step1:

Given function F(x) = (x² - 1)(x² - 4) and closed interval [a, b] = [-3, 3].

Write the above equation in general form of polynomial function P(x) = a_nxⁿ + a_n-1x^n-1 + . . . + a₁x + a₀.

Apply Distributive property: a(b - c) = ab - ac.

F(x) = x²(x² - 4) - 1(x² - 4)

F(x) = x⁴ - 4x² - x² + 4.

F(x) = x⁴ - 5x² + 4.

(1) f(x) is continuous on closed interval [-3, 3] where -3 < 3.

(2) f(x) is differentiable on the open interval (-3, 3).

F(a) = F(-3) = (-3)⁴ - 5(-3)² + 4 = 81 - 45 + 4 = 40.

F(b) = F(3) = (3)⁴ - 5(3)² + 4 = 81 - 45 + 4 = 40.

Step2:

Condition 1 and 2 are satisfied

So there exist a point c in the (-3, 3) such that .

Substitute a = -3, b = 3, F(b) = 40 and F(a) = 40 in the above formula.

F'(c) = (40 - 40)/(3 + 3) = 0.

Differentiate the function F(x) with respect to x.

F'(x) = 4x³ - 10x

Therefore, 4x³ - 10x = 0.

2x(2x² - 5) = 0

2x = 0 and 2x² - 5 = 0

x = 0 and x = ± √(5/2).

The values of c = 0 and c = ± √(5/2) and these are lies in the interval [-3, 3].