Question

Verify Rolle's theorem for the function $f\left(x\right)=x^2+2x-8,xϵ[-4,2]$

Answer 1

Given function is $f\left(x\right)=x^2+2x-8,xϵ[-4,2]$

Since, a polynomial function is continous and derivable on R therefore

(i) f(x) is continuous on (-4,2)

(ii) f(x) is derivable on (-4,2)

Also, f(-4)=$(-4{)}^2$+(-4)-8=0 $(\therefore f(x)=x^2+2x-8)$

and f(2)=$2^2+2\times 2-8=0$ $\Rightarrow f(-4)=f\left(2\right)$

This means that all the conditions of Rolle's theorem are satisfied by f(x) in [-4,2]

Therefore, it exists atleast one real $xϵ[-4,2]$ such that f(c)=0.

Now, $f\left(x\right)=x^2+2x-8\Rightarrow f^{\prime }\left(x\right)\frac{d}{dx}(x^2+2x-8)=2x+2$

Putting $f^{\prime }\left(c\right)=0\Rightarrow 2c+2=0\Rightarrow c=-1$.

Thus, f(-1)=0 and -1 $ϵ[-4,2]$.

Rolle's theorem is varified with c=-1.