Question
Verify Rolle's theorem of the following function on the indicated interval
$\text{f}(\text{x})=4^{\sin\text{x}}\text{ on }[0,\pi]$
$\text{f}(\text{x})=4^{\sin\text{x}}\text{ on }[0,\pi]$
We know that exponential and
$\sin\text{x}$ both are continuous and differentiable, so f(x) is continuous is $[0,\pi]$ and differentiable is $(0,\pi).$Now,
$\text{f}(0)=4^{\sin0}=4^0=1$ $\text{f}(\pi)=4^{\sin\pi}=4^0=1$ $\Rightarrow\text{f}(0)=\text{f}(\pi)$ So, Rolle's theorem is applicable, there must exist a point $\text{c}\in(0,\pi)$ such that f'(c) = 0. Now, $\text{f}(\text{x})=4^{\sin\text{x}}$ $\text{f}'(\text{x})=4^{\sin\text{x}}\log4\times\cos\text{x}$ Now, $\text{f}'(\text{c})=0$ $4^{\sin\text{c}}\times\cos\times\text{c}\log4=0$ $\Rightarrow\cos\text{c}=0$ $\Rightarrow\text{c}=\frac{\pi}{2}\in(0,\pi)$ Hence, Rolle's theorem is verified.Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$(1+\text{x})(1+\text{y}^2)\text{dx}+(1+\text{y})(1+\text{x}^2)\text{dy}=0$
$(1+\text{y}^2)\tan^{-1}\text{xdx}+2\text{y}(1+\text{x}^2)\text{dy}=0$