Appropriate Question :-
If f(x) = tanx, show that f(0) = f(π). Is Rolle's theorem applicable to f(x) in [0, π]? Give reason.
[tex]\large\underline{\sf{Solution-}}[/tex]
Given function is
[tex]\sf \: f(x) = tanx \\ \\ [/tex]
So, Consider
[tex]\sf \: f(0) = tan0 = 0 \\ \\ [/tex]
Now, Consider
[tex]\sf \: f(\pi) = tan\pi = 0 \\ \\ [/tex]
Thus,
[tex]\bf\implies \:f(0) = f(\pi) \\ \\ [/tex]
Now, we have
[tex]\sf \: f(x) = tanx ,\: \: x \in \: [0, \: \pi] \\ \\ [/tex]
We know,
[tex]\sf \: tan\dfrac{\pi}{2} =not \: defined \\ \\ [/tex]
[tex]\bf\implies \:tanx \: is \: not \: continuous \: at \: x = \dfrac{\pi}{2} \\ \\ [/tex]
[tex]\bf\implies \:Rolle's \: theorem \: is \: not \: applicable \: \\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Basic Concept Used :-
Let f(x) be a real valued function defined on [a, b] such that
1. f(x) is continuous on [a, b].
2. f(x) is differentiable on (a, b).
3. f(a) = f(b)
Then, Rolle's Theorem is applicable.
Therefore, there exist atleast one real number c belongs to (a, b) such that f'(c) = 0
[tex]\rule{190pt}{2pt}[/tex]
Additional Information :-
[tex]\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {x}^{n} & \sf {nx}^{n - 1}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}[/tex]