Maxima and Minima — Maths STD 12 Science — Question

3. $\frac{-1}{\text{e}}$

Solution :

Here, $\text{f}(\text{x})=\text{x}\log_{\text{e}}\text{x}$

lmplies that $\text{f}'(\text{x})=\log_{\text{e}}\text{x}+1$

For a local maxima or a local minima, we must have f'(x) = 0

lmplies that $\log_{\text{e}}\text{x}+1=0$ 

lmplies that $\log_\text{e}\text{x}=-1$

lmplies that $\text{x}=\text{e}^{-1}$

Now, $\text{f}''(\text{x})=\frac{1}{\text{x}}$

lmplies that $\text{f}''(\text{e}^{-1})=\text{e}>0$

Threrfore, $\text{f}(\text{e}^{-1})$ is a local minima.

Hence, the minimum value of $\text{f}(\text{x})=\text{f}(\text{e}^{-1})$

lmplies that $\text{e}^{-1}\log_\text{e}(\text{e}^{-1})=-\text{e}^{-1}=\frac{-1}{\text{e}}$