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]?

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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]?

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Answer 1 · 2018-04-14T08:11:58

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 $x=2$

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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]?

1 Answer

Explanation:

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Science

Math

Humanities

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

1 Answer

Explanation:

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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]?