this is section 3.7 problem 6: for the function $h(x)=-x^{3}-3x^{2}+15x + 3$, determine the absolute maximum…

Explanation:

Step1: Find the derivative

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

Step2: Set the derivative equal to zero

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

Step3: Calculate function values at critical points and endpoints

Calculate $h(x)$ at $x=-6$

$h(-6)=-(-6)^{3}-3(-6)^{2}+15(-6)+3=216-108 - 90+3=21$

Calculate $h(x)$ at $x=-1-\sqrt{6}\approx - 3.45$

$h(-1 - \sqrt{6})=-(-1-\sqrt{6})^{3}-3(-1-\sqrt{6})^{2}+15(-1-\sqrt{6})+3$ $=-(-1 - 3\sqrt{6}-18 - 6\sqrt{6})-3(1 + 2\sqrt{6}+6)-15 - 15\sqrt{6}+3$ $=(19 + 9\sqrt{6})-(21 + 6\sqrt{6})-15 - 15\sqrt{6}+3$ $=19 + 9\sqrt{6}-21 - 6\sqrt{6}-15 - 15\sqrt{6}+3=-14 - 12\sqrt{6}\approx - 43.4$

Calculate $h(x)$ at $x=-1+\sqrt{6}\approx1.45$

$h(-1+\sqrt{6})=-(-1+\sqrt{6})^{3}-3(-1+\sqrt{6})^{2}+15(-1+\sqrt{6})+3$ $=-(-1 + 3\sqrt{6}-18+6\sqrt{6})-3(1 - 2\sqrt{6}+6)-15 + 15\sqrt{6}+3$ $=(19 - 9\sqrt{6})-(21 - 6\sqrt{6})-15 + 15\sqrt{6}+3=-14 + 12\sqrt{6}\approx19.9$

Calculate $h(x)$ at $x = 3$

$h(3)=-(3)^{3}-3(3)^{2}+15(3)+3=-27-27 + 45+3=-6$

Answer:

Absolute maximum = 21 at $x=-6$ Absolute minimum = - 43.4 at $x=-1-\sqrt{6}$