Question

Let $f(x)=\int\limits_0^x 3^t\left(3^t-4\right)(x-t) d t \quad(x \geq 0)$. If $x=a$ is the point where $f(x)$ attains its local minimum value then find the value of $3^{ a }$.

Answer 1

Given, $ f(x)=x \int\limits_0^x 3^t\left(3^t-4\right) d t-\int\limits_0^x 3^t \cdot t\left(3^t-4\right) d t$
For maxima $/ \operatorname{minima}$, we have
$ f ^{\prime}( x )=0$
$\Rightarrow \int\limits_0^x 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) $
$\therefore f ^{\prime}( x )=\int\limits_0^{ x } 3^{ t }\left(3^{ t }-4\right) dt$
$f ^{\prime}( x )=\frac{1}{2 \ln 3}\left(3^{2 x }-8 \cdot 3^{ x }+7\right)$
$ 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$
$x =\log _3 7 \text { is the point of minima. }$
$\text { Hence, } 3^{ a }=3^{\log _3 7}=7$