#### Provide solution for R.D.Sharma class 12 chapter 17 Maxima and Minima Excercise 17.3 question 1 sub Question 5.

Point of local minima value is x=-1 and it’s local minimum value is $-\frac{1}{e^{.}}$

Hint:

First find critical values of f(x) by solving f'(x) =0 then find f''(x).

First find critical values of f(x) by solving f'(x) =0 then find f''(x).

If $f"(c_1) > 0$ then $c_1$ is point of local minima.

If $f"(c_2) < 0$ then $c_2$ is point of local maxima .

where $c_1$ & $c_2$ are critical points.

Put $c_1$ and $c_2$ in f(x) to get minimum value & maximum value.

Given:$f(x)=x e^{\mathrm{x}}$

Explanation:

We have,

\begin{aligned} &f(x)=x e^{\mathrm{x}} \\ &f^{\prime}(x)=e^{\mathrm{x}}(x+1) \ldots \text { using } \frac{d}{d x} u . v=\frac{u d v}{d x}+\frac{v d u}{d x} \\ &f^{\prime}(x)=e^{\mathrm{x}}(x+1) \\ &f^{\prime \prime}(x)=e^{\mathrm{x}}(x+1)+e^{\mathrm{x}} \ldots u \sin g \frac{d}{d x} u . v=\frac{u d v}{d x}+\frac{v d u}{d x} \\ &f^{\prime \prime}(x)=e^{\mathrm{x}}(x+2) \end{aligned}

Now, for maxima & minima, f’(x)=0

\begin{aligned} e^{\mathrm{x}}(x+1) &=0 \\ x+1 &=0 \\ x &=-1 \end{aligned}
at x=-1,

$f^{\prime \prime}(-1)=e^{-1}(-1+2)=\frac{1}{e}$

X = -1 is point of local minima.

So, local min. value at x= 1 is

$f(-1)=-1 e^{-1}=-1 / e$

Thus, point of local minima is -1 & its min value is $-\frac{1}{e}$