Answer to Question #89446 in Calculus for adigam amos jacob

Question

let f(x)=x^4-2x^2. Find the all c (where c is the interception on the x-axis) in the interval (-2,2) such that f'(x)=0 using Rolle's theorem.

Answer 1

Solution. Using Rolle's theorem. Suppose that a function f(x)  is continuous on the closed interval [a;b] and differentiable on the open interval (a;b). Then if f(a)=f(b), then there exists at least one point c in the open interval (a;b) for which f'(c)=0.

Function  f(x)=x^4-2x^2  is continuous on the closed interval [-2;2] and differentiable on the open interval (-2;2).

$$f(-2)=(-2)^4-2*(-2)^2=16-8=8$$

$$f(2)=2^4-2*2^2=16-8=8$$ $$f(-2)=f(2)$$

On the other hand

$$f'(x)=4x^3-4x.$$

Hence get equation

$$4x^3-4x=0$$

Solve the equation

$$4x(x^2-1)=0$$

Roots of the equation

$$x_1=0 \isin (-2;2)$$

$$x_2=-1 \isin (-2;2)$$

$$x_3=1 \isin (-2;2)$$

Answer. -1, 0, 1.