If `f:[-5,\ 5]->R` is differentiable and if `f^(prime)(x)` doesnt vanish anywhere, then prove that `f(-5)!=f(5)` .

Explanation

As given f :`[-5,5]` → R is a differentiable function.

Since every differentiable function is a continuous function, we obtain

(a) f is continuous on that`[−5, 5].`

(b) f is differentiable on that `(−5, 5).`

Hence, by the Mean Value Theorem, there exists c `∈(−5, 5)` such that



`f^{\prime}(c)=\frac{f(5)-f(-5)}{5-(-5)}`

...