Find the absolute maximum and absolute minimum values of f on the given interval. f(x) = ln(x2 + 2x + 4), [βˆ’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) =

Accepted Solution

[tex]f(x)=\ln(x^2+2x+4)\implies f'(x)=\dfrac{2x+2}{x^2+2x+4}[/tex]The numerator determines where the derivative vanishes (the denominator has a minimum value of 3, since [tex]x^2+2x+4=(x+1)^2+3\ge3[/tex]).[tex]2x+2=0\implies x=-1[/tex]At this critical point, we have[tex]f(-1)=\ln((-1)^2+2(-1)+4)=\ln3\approx1.099[/tex]At the endpoints, we have[tex]f(-2)=\ln4\approx1.386[/tex][tex]f(2)=\ln12\approx2.485[/tex]so [tex]f[/tex] attains a maximum value of [tex]\ln12[/tex] and a minimum value of [tex]\ln3[/tex].