Question

How do you find the absolute maximum and absolute minimum values of f on the given interval `f(x) = ln(x^2 + 3x + 9)` and [-2, 2]?

Answer 1

• Minimum : `ln(27/4)` at `x=-3/2`
• Maximum : `ln(19)` at `x=2`

Explanation:

Given

`f(x) = ln(x^2 + 3x + 9)`

we have

`f^'(x) = (2x+3)/(x^2 + 3x + 9)`

and

`f^{' '}(x) = ((x^2 + 3x + 9)d/dx(2x+3)-(2x+3)d/dx(x^2 + 3x + 9))/(x^2 + 3x + 9)^2`
`qquad = ((x^2 + 3x + 9)*2-(2x+3)^2)/(x^2 + 3x + 9)^2`
`qquad = (2x^2+6x+18-(4x^2+12x+9))/(x^2 + 3x + 9)^2`
`qquad = (9-2x^2-6x)/(x^2 + 3x + 9)^2`

For a local extremum, we have

`f^'(x)=0 implies 2x+3 = 0 implies x= -3/2`

`f^{''}(-3/2) >0`

So `x=-3/2` is a local minimum. The minimum value is

`f(-3/2) =ln((-3/2)^2+3(-3/2)+9)=ln(27/4)`

Since `x=-3/2` is the only local extremum in `[-2,2]`, the maximum is reached at an endpoint.

Now,

`f(-2) = ln ((-2)^2+3(-2)+9) = ln(7)`

and

`f(2) = ln (2^2+3*2+9) = ln(19)`

So the global maximum is `ln(19)` at `x=2`