Answer to Question #114616 in Calculus for ANJU JAYACHANDRAN

Question

a)Let f(x)=tan x
i)Show that there exists no c in the interval ]0, pi[such that f'(c)=0 even though f(0)=f(pi)=0
ii)Explain why the result in part (a) does not violate Rolle's Theorem.

Answer 1

a) $f'(x) = sec^2 x \neq 0$ at any $c \in (0,\pi)$, even through $f(0)=f(\pi)=0.$

b) Rolle's Theorem has three hypotheses:

f is continuous on the closed interval [a,b].

f is differentiable on the open interval (a,b).

f(a)=f(b).

Then there exist at least one $c \in (a,b)$ such that $f'(c)=0.$

Now given function f(x) = tan(x) is not continuous on $[0,\pi ],$ because $f(x)= tan(x)$ is not defined at $x = \pi/2.$

And The derivative function sec²(x) is not defined at $\pi/2$, hence the given finction is not diferentiable in $(0,\pi).$

And $f(0)=f(\pi)=0.$

Thus, results are not contradictory to Rolle's theorem. Rolle's theorem violates because in this problem Rolle's theorem hypotheses do not hold.