Question

Define the values of a and b of f(x)= 1/(x^2+ax+b) so that you get a maximum in point (3/2,-4). Could you help me to find the answer?

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 can get to the answer?

Answer 1

`a=-3, b=2`

`f(x) = 1/(x^2-3x+2)`

Explanation:

We have:

`f(x) = 1/(x^2+ax+b)`
`" " = (x^2+ax+b)^(-1)`

Firstly, we not that:

`f(3/2) = -4 => 1/(9/4+3/2a+b) = -4`
`:. 1 = -4( 9/4+3/2a+b)`
`:. -1 = 9+6a+4b`
`:. 3a+2b = -5` ..... [A}

Differentiating wrt `x` we get:

`f'(x) = (-1)(x^2+ax+b)^(-2)(2x+a)`
`" " = - (2x+a)/(x^2+ax+b)^2`

At any maximum or minimum we must have:

`f'(x)=0 => - (2x+a)/(x^2+ax+b)^2 = 0`
`:. 2x+a = 0 => x=-a/2`

As the given coordinate `(3/2,-4)` is to be a maximum then:

`3/2 = -a/2 => a=-3`

Substituting `a=-3` into [A] we get:

`-9 +2b = -5 => 2b=4 => b=2`

Hence, we have:

`f(x) = 1/(x^2-3x+2)`

And the associated graph is:
graph{1/(x^2-3x+2) [-3, 5, -10, 10]}

It certainly appears as if the graphs fits the given criteria, and if required we could perform a second derivative test to establish that `f(3/2) lt 0`