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Question

The maximum value of the function f(x)=2x315x2+36x48 on the set A={x|x2+209x} is

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Solution

x²+209x

x²9x+200

x²4x5x+200

x(x4)5(x4)0

(x4)(x5)0

4x5

A={4x5}

now,
f(x)=2x³15x²+36x48

differentiate wrt x

f(x)=6x²30x+36=6(x²5x+6)

=6(x2)(x3)

So f(x) is increasing in (,2)(3,)

maximum value of f(x) at x=5

f(5)=2(5)³15(5)²+36(5)48

=250375+18048

=430423=7

fmax=f(5)=7

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