Question

Find the 'c' value, that proves Rolle's Theorem true, if possible. f(x) = tanx [0, π] ?

Answer 1

Kindly refer to the Discussion in Explanation.

Explanation:

Here is Rolle's Theorem :

Suppose that a function (fun.) `f : [a,b] to RR" is,"`

(R_1) : continuous on `[a,b];`

(R_2) : derivable on `(a,b);`

(R_3) : `f(a)=f(b).`

Then, `EE" a "c in (a,b)" such that "f'(c)=0.`

In our case, `f(x)=tanx, a=0, b=pi.`

Note that, `f(x)=tanx,` is not defined at `x=pi/2 in [0,pi].`

Hence, `f` is not continuous on `(o,pi).`

In other words, the condition (R_1) is violated by `f`.

Accordingly, we can not apply Rolle's Theorem to `f=tan.`

Even otherwise, `AA x in RR-{2npi+-pi/2 | n in ZZ}, |secx| ge 1.`

So, `f'(x)=sec^2x !=0, AA x in RR-{2npi+-pi/2 | n in ZZ}.`

Enjoy Maths.!