this is similar to section 3.7 problem 16: for the function h(x)=-x^3 - 3x^2+15x+(1), determine the absolute…

Explanation:

Step1: Find the derivative

$h(x)=-x^{3}-3x^{2}+15x + 1$, then $h'(x)=-3x^{2}-6x + 15$.

Step2: Set the derivative equal to 0

$-3x^{2}-6x + 15=0$. Divide through by - 3: $x^{2}+2x - 5=0$. Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ with $a = 1$, $b = 2$ and $c=-5$, we have $x=\frac{-2\pm\sqrt{4+20}}{2}=\frac{-2\pm\sqrt{24}}{2}=-1\pm\sqrt{6}$.

Step3: Consider values in the interval [0,2]

We only consider $x=-1 + \sqrt{6}\approx - 1+2.45=1.45$ since $-1-\sqrt{6}\approx-1 - 2.45=-3.45$ is outside the interval $[0,2]$.

Step4: Evaluate the function at critical and end - points

$h(0)=-0^{3}-3\times0^{2}+15\times0 + 1=1$. $h(1.45)=-(1.45)^{3}-3\times(1.45)^{2}+15\times1.45 + 1=-3.05 - 6.30+21.75 + 1=13.40$. $h(2)=-2^{3}-3\times2^{2}+15\times2 + 1=-8-12 + 30+1=11$.

Answer:

Absolute maximum = 13.40 at $x = 1.45$. Absolute minimum = 1.00 at $x = 0$.