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

Answers (1)

Answer:

              \left ( 3,19 \right )

Hint:

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

Given:

 f\left ( x \right )=\left ( x-1 \right )^{2}

Solution:

f\left ( x \right ) is the sum of a square and a constant term. It will acquire minimum value when the square term becomes zero

Hence

f\left ( x \right ) is max at x=-3

f\left ( -3 \right )=19

Or

\begin{aligned} &f^{\prime}(x)=(x-1)^{2}+1 \\ &f^{\prime}(x)=2(x-1)=0 \\ &x \in[-3,1] \end{aligned}

\begin{aligned} &f^{\prime}(x)=2> 0 \end{aligned}

x=1 is local minimum

\begin{aligned} &f^{\prime \prime}(x)=2>0\\ &x=1 \text { is local minimum }\\ &f(-3)=(-3-1)^{2}+3\\ &=19(m, M)\\ &=(3,19)\\ &f(1)=3 \rightarrow \min =m \end{aligned}

 

 

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