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Please solve RD Sharma class 12 chapter Maxima and Minima exercise 17.4 question 5 maths textbook solution

Answers (1)

Answer:

        Absolute maximum =56 \text { at } x=5

        Absolute minimum =24 \text { at } x=1

Hint:

        check the value of f(x) at end points & where f ‘(x) = 0

Given:

        f(x)=2 x^{3}-15 x^{2}+36 x+1 \text { on }[1,5]

Explanation:

        f^{\prime}(x)=6 x^{2}-30 x+36

        \begin{aligned} &f^{\prime}(x)=0 \\ &6 x^{2}-30 x+36=0 \\ &x^{2}-5 x+6=0 \\ &x^{2}-3 x-2 x+6=0 \\ &x(x-3)-2(x-3)=0 \end{aligned}

        \begin{aligned} &(x-2)(x-3)=0 \\ &x-2=0, x-3=0 \\ &x=2,3 \end{aligned}

Check the value of f(x) \text { at } 1,2,3,5

        \begin{aligned} &f(1)=2 \times 1^{3}-15 \times 1^{2}+36 \times 1+1=24 \\ &f(2)=2 \times 2^{3}-15 \times 2^{2}+36 \times 2+1=29 \\ &f(3)=2 \times 3^{3}-15 \times 3^{2}+36 \times 3+1=28 \\ &f(5)=2 \times 5^{3}-15 \times 5^{2}+36 \times 5+1=56 \end{aligned}

Hence,

Absolute maximum =56 \text { at } x=5

Absolute minimum =24 \text { at } x=1

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