Question

Let $f(x)=\underset{0}{\overset{x}{\int }}{3}^t\left(3^t-4\right)(x-t)dt(x\ge 0)$. If $x=a$ is the point where $f(x)$ attains its local minimum value then find the value of $3^a$.

Answer 1

Answer: 7

Given, $f(x)=x\underset{0}{\overset{x}{\int }}{3}^t\left(3^t-4\right)dt-\underset{0}{\overset{x}{\int }}{3}^t\cdot t\left(3^t-4\right)dt$
For maxima $/\mathrm{minima}$, we have
$f^{\prime }(x)=0$
$\Rightarrow \underset{0}{\overset{x}{\int }}{3}^t\left(3^t-4\right)dt+x\cdot 3^x\left(3^x-4\right)-3^x\cdot x\left(3^x-4\right)$
$\therefore f^{\prime }(x)=\underset{0}{\overset{x}{\int }}{3}^t\left(3^t-4\right)dt$
$f^{\prime }(x)=\frac{1}{2\ln 3}\left(3^{2x}-8\cdot 3^x+7\right)$
$f^{\prime }(x)=\frac{1}{2\ln 3}\left(3^x-1\right)\left(3^x-7\right)$

$f^{\prime }(x)=0\Rightarrow x=0,{\log }_{3}7$
$x={\log }_{3}7\text{ is the point of minima. }$
$\text{ Hence, }{3}^a=3^{{\log }_{3}7}=7$