this is similar to section 3.7 problem 36: for the function h(x)=-x^3 - 3x^2+15x+(5), determine the absolute…

Explanation:

Step1: Find the derivative

Differentiate $h(x)=-x^{3}-3x^{2}+15x + 5$ using power - rule. $h'(x)=-3x^{2}-6x + 15$.

Step2: Set the derivative equal to 0

Solve $-3x^{2}-6x + 15=0$. Divide through by -3: $x^{2}+2x - 5=0$. Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ with $a = 1$, $b = 2$, $c=-5$. $x=\frac{-2\pm\sqrt{2^{2}-4\times1\times(-5)}}{2\times1}=\frac{-2\pm\sqrt{4 + 20}}{2}=\frac{-2\pm\sqrt{24}}{2}=\frac{-2\pm2\sqrt{6}}{2}=-1\pm\sqrt{6}$. We consider only the value in the interval $(0,4)$. So $x=-1+\sqrt{6}\approx - 1+2.45 = 1.45$.

Step3: Evaluate the function at critical point and endpoints

Evaluate $h(x)$ at $x = 1.45$, $x = 0$ (limit as we approach from the right since it's an open - interval), and $x = 4$. $h(1.45)=-(1.45)^{3}-3(1.45)^{2}+15(1.45)+5=-3.05 - 6.30+21.75 + 5=17.40$. $\lim_{x\rightarrow0^{+}}h(x)=5$. $h(4)=-(4)^{3}-3(4)^{2}+15(4)+5=-64-48 + 60+5=-47$.

Answer:

Absolute maximum = $17.40$ at $x = 1.45$. Absolute minimum = $-47$