Question

verify Rolle's theorem for f on [-1, 1] defined by (x) =x^4 -4x^2 +7

Answer 1

$f(x)=x^4-4x^2+7$ is continuous on $[-1,1]$ as a polynomial.

$f(x)=x^4-4x^2+7$ is differentiable on $(-1,1)$ as a polynomial.

$$f(-1)=(-1)^4-4(-1)^2+7=4$$

$$f(1)=(1)^4-4(1)^2+7=4$$

$$f(-1)=4=f(1)$$

Since the function $f(x)=x^4-4x^2+7$ satisfies these conditions, then the function $f(x)=x^4-4x^2+7$ satisfies the Rolle's theorem.

Then ther is the number $c$ in $(-1, 1)$ such that $f'(c)=0.$

$$f'(x)=(x^4-4x^2+7)'=4x^3-8x$$

$$f'(x)=0=>4x^3-8x=0$$

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

$$x_1=0, x_2=-\sqrt{2}, x_3=\sqrt{2}$$

Since the function $f(x)=x^4-4x^2+7$ is defined on $[-1, 1],$ then $c=0$ and

$$f'(c)=f'(0)=0.$$