Question

Find the absolute maximum and absolute minimum values of f on the given interval. f(x) = ln(x2 + 2x + 4), [−2, 2] Step 1 The absolute maximum and absolute minimum values of the function f occur either at a critical number or at an endpoint of the interval. Recall that a critical number is a value of x where f '(x) = 0 or where f '(x) doesn't exist. We begin by finding the critical numbers. f '(x) =

Answer 1

$f(x)=\ln(x^2+2x+4)\implies f'(x)=\dfrac{2x+2}{x^2+2x+4}$

The numerator determines where the derivative vanishes (the denominator has a minimum value of 3, since $x^2+2x+4=(x+1)^2+3\ge3$).

$2x+2=0\implies x=-1$

At this critical point, we have

$f(-1)=\ln((-1)^2+2(-1)+4)=\ln3\approx1.099$

At the endpoints, we have

$f(-2)=\ln4\approx1.386$

$f(2)=\ln12\approx2.485$

so $f$ attains a maximum value of $\ln12$ and a minimum value of $\ln3$.