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  • Need solution for R.D.Sharma maths class 12 chapter Maxima and Minima exercise 17.3 Question 1 sub Question 6.

Need solution for R.D.Sharma maths class 12 chapter  Maxima and Minima exercise 17.3 Question 1 sub Question 6.

Answers (1)

Answer:

Point of local minima value is 2 and it’s local minimum value is 2.

Hint:

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

If f^{\prime \prime}\left(c_{1}\right)>0 then \mathrm{c}_{1} is point of local minima.

If f^{\prime \prime}\left(c_{2}\right)>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:

\frac{x}{2}+\frac{2}{x}, \underline{x}>0

Explanation:

We have,

\begin{aligned} &f(x)=\frac{x}{2}+\frac{2}{x} \quad, x>0 \\ &f^{\prime}(x)=\frac{1}{2}-\frac{2}{x^{2}} \ \\ &f^{\prime \prime}(x)=\frac{4}{x^{3}} \end{aligned}

For local maxima or minima, we have f^{\prime}(x)=0

\begin{gathered} \frac{1}{2}-\frac{2}{x^{2}}=0 \\ x^{2}=4 \\ x=2 \text { or } x=-2 \end{gathered}

    since, x>0

    x = 2

Now, at x=2 ,

f(2)=\frac{2}{2}+\frac{2}{2}=2

 Thus, its min. value at 2 is 2.

 

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