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Explain Solution for RD Sharma Maths Class 12 Chapter 17 Maxima and Minima Exercise Fill in the blanks Question 9 maths textbook solution

Answers (1)

Answer:

              \sqrt{\frac{b}{a}}

Hint:

For maxima or minima we must have {f}'\left ( x \right )=0

Given:

\begin{aligned} &f(x)=a x+\frac{b}{x} \\ \end{aligned}

Solution:

\begin{aligned} &f(x)=a x+\frac{b}{x} \\ &f^{\prime}(x)=a-\frac{b}{x^{2}} \\ &f^{\prime}(x)=0 \\ &a=\frac{b}{x^{2}} \text { or } x=\sqrt{\frac{b}{a}} \end{aligned}

Now

f^{\prime \prime}(x)=\frac{2 b}{x^{3}}>0 \text { for } x=\sqrt{\frac{b}{a}}  

Thus least value of f\left ( x \right )  is x=\sqrt{\frac{b}{a}}

 

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