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 $a=-1$. Find the value of $b$.

Answer 1

$f\left(x\right)=x^4+x^2-2$

Since the hypothesis of Rolle's theorem are satisfied by $f$ in the interval $[a,b]$, we have

$f\left(a\right)=f\left(b\right)$, where $a=-1$

Now, $f\left(a\right)$

$=f(-1)$

$=(-1)4+(-1)2-2$

$=1+1-2$

$=0$

and $f\left(b\right)$

$=b^4+b^2-2$

$\therefore f\left(a\right)=f\left(b\right)$ gives

$0=b^4+b^2-2$

i.e. $b^4+b^2-2=0$

Since, $b=1$ satisfies this equation, $b=1$ is one of the root of this equation.

Hence, $b=1$.