Question

use a table to graph (f(x)=-2x^{3}+12x^{2}-8x) on a separate sheet of paper. then estimate the (x)-coordinates at which relative maxima and relative minima occur.
the relative maximum occurs between the integers (x=) and (x=)
the relative minimum occurs between the integers (x=) and (x=)

Explanation:

Step1: Find the derivative

Differentiate $f(x)=-2x^{3}+12x^{2}-8x$ using the power - rule. The derivative $f'(x)=-6x^{2}+24x - 8$.

Step2: Set the derivative equal to zero

Solve the quadratic equation $-6x^{2}+24x - 8 = 0$. Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for $ax^{2}+bx + c = 0$, here $a=-6$, $b = 24$, $c=-8$. So $x=\frac{-24\pm\sqrt{24^{2}-4\times(-6)\times(-8)}}{2\times(-6)}=\frac{-24\pm\sqrt{576 - 192}}{-12}=\frac{-24\pm\sqrt{384}}{-12}=\frac{-24\pm8\sqrt{6}}{-12}=2\pm\frac{2\sqrt{6}}{3}\approx2\pm1.63$.

Step3: Determine the intervals

The critical points are $x_1=2-\frac{2\sqrt{6}}{3}\approx0.37$ and $x_2=2 + \frac{2\sqrt{6}}{3}\approx3.63$. We can test the sign of $f'(x)$ in the intervals $(-\infty,x_1)$, $(x_1,x_2)$ and $(x_2,\infty)$.

Step4: Identify maxima and minima

$f'(x)$ changes sign from positive to negative at $x = 2+\frac{2\sqrt{6}}{3}\approx3.63$ (relative maximum) and from negative to positive at $x=2-\frac{2\sqrt{6}}{3}\approx0.37$ (relative minimum).

Answer:

The relative maximum occurs between the integers $x = 3$ and $x = 4$.
The relative minimum occurs between the integers $x = 0$ and $x = 1$.