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

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

Answers (1)

Answer:

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}

 

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