Question

Given an interval $[a,b]$ that satisfies hypothesis of Rolle's theorem for the function $f\left(x\right)=x^4+x^2-2.$ It is known that $a=-1.$ Then the possible value of $b$ is :

Answer 1

We factor the polynomial:
$x^4+x^2-2=(x^2+2)(x^2-1)=(x^2+2)(x-1)(x+1).$
It is now easy to see that the function has two zeros: $x_1=-1($ coincides with the value of $a)$ and $x_2=1.$
Since the function is a polynomial, it is everywhere continuous and differentiable. So this function satisfies Rolle's theorem on the interval $[-1,1].$ Hence, $b=1$