Define the values of a and b so that you get a maximum in point (3/2,-4). Could you help me to find the answer?

Question

You have to find a and b that are an element of f(x) and f(x) = #1/(x^2+ax+b)#. And it has to have a maximum in point
( #3/2 ,-4#) . Can you explain in detail how you've got the answer?

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Answer 1 · 2018-01-09T18:21:57

# a=-3, and b=2#.

Given that, #f_max=f(3/2)=-4#. But, #f(x)=1/(x^2+ax+b)#.

Hence, #f(3/2)=-4 rArr -4=1/{(3/2)^2+a(3/2)+b#.

#rArr -4=4/(9+6a+4b), or, 3a+2b=-5..................(ast)#.

Further, for #f_max, f'(x)=0, f''(x)lt0#.

Since, #f_max=f(3/2), :. f'(3/2)=0, and f''(3/2)lt 0#.

#f(x)=1/(x^2+ax+b)#

#rArr f'(x)=-1/(x^2+ax+b)^2*(2x+a)#.

#:. f'(3/2)=0 rArr 2(3/2)+a=0 rArr a=-3#.

#a=-3 and (ast) rArr 3(-3)+2b=-5 rArr b=2#