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  • Explain Solution for RD Sharma Maths Class 12 Chapter 17 Maxima and Minima Exercise Fill in the blanks Question 1 maths textbook solution

Explain Solution for RD Sharma Maths Class 12 Chapter 17 Maxima and Minima Exercise Fill in the blanks Question 1 maths textbook solution

Answers (1)

Answer:

              \frac{4}{3}

Hint:

For maxima or minima we must have {f}'(x)

Given:

f(x)=\frac{1}{4 x^{2}+2 x+1}

Solution:

We have

f(x)=\frac{1}{4 x^{2}+2 x+1}

{f}'(x)=\frac{-\left ( 8x+2 \right )}{\left ( 4 x^{2}+2 x+1 \right )^{2}}

Put  {f}'(x)=0

8x+2=0

x=-\frac{1}{4}

f^{\prime \prime}(x)=\frac{\left[-\left(4 x^{2}+2 x+1\right)^{2} \times 8-(8 x+2) \times 2 \times\left(4 x^{2}+2 x+1\right)(8 x+2)\right]}{\left[4 x^{2}+2 x+1\right]^{4}}  

Now f^{\prime \prime}(\frac{-1}{4})  is negative (point of maxima)

So

\begin{aligned} &f\left(\frac{-1}{4}\right)=\frac{1}{\left(4 \times\left(\frac{1}{16}\right)-2 \times\left(\frac{1}{4}\right)+1\right)} \\ &f\left(\frac{-1}{4}\right)=\frac{4}{3} \end{aligned}

 

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