the polynomial function h is defined by h(x)=-3x^3 + 9x^2 - 3x - 4. use the aleks graphing calculator to…

Answer:

Since we are required to use the ALEKS graphing - calculator, this response will assume the use of the calculator's functionality. Without actually using the ALEKS graphing calculator, we can also use calculus. First, find the derivative of the function $h(x)=-3x^{3}+9x^{2}-3x - 4$.

The derivative $h'(x)$ is given by: [ \begin{align*} h'(x)&=\frac{d}{dx}(-3x^{3}+9x^{2}-3x - 4)\ &=-9x^{2}+18x - 3 \end{align*} ]

Set $h'(x) = 0$ to find the critical points: [ -9x^{2}+18x - 3=0 ] Divide through by $- 3$: [3x^{2}-6x + 1=0]

Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $ax^{2}+bx + c = 0$, here $a = 3$, $b=-6$, and $c = 1$.

[ \begin{align*} x&=\frac{6\pm\sqrt{(-6)^{2}-4\times3\times1}}{2\times3}\ &=\frac{6\pm\sqrt{36 - 12}}{6}\ &=\frac{6\pm\sqrt{24}}{6}\ &=\frac{6\pm2\sqrt{6}}{6}\ & = 1\pm\frac{\sqrt{6}}{3} \end{align*} ]

[x_1=1+\frac{\sqrt{6}}{3}\approx1 + 0.82=1.82] [x_2=1-\frac{\sqrt{6}}{3}\approx1-0.82 = 0.18]

Now, find the second - derivative $h''(x)$: [ h''(x)=\frac{d}{dx}(-9x^{2}+18x - 3)=-18x + 18 ]

Evaluate $h''(x)$ at the critical points:

For $x = 1+\frac{\sqrt{6}}{3}$: [ \begin{align*} h''(1+\frac{\sqrt{6}}{3})&=-18(1+\frac{\sqrt{6}}{3})+18\ &=-18 - 6\sqrt{6}+18\ &=-6\sqrt{6}<0 \end{align*} ]

For $x = 1-\frac{\sqrt{6}}{3}$: [ \begin{align*} h''(1-\frac{\sqrt{6}}{3})&=-18(1-\frac{\sqrt{6}}{3})+18\ &=-18 + 6\sqrt{6}+18\ &=6\sqrt{6}>0 \end{align*} ]

Since $h''(1-\frac{\sqrt{6}}{3})>0$, the function has a local minimum at $x = 1-\frac{\sqrt{6}}{3}\approx0.18$.

Find $h(0.18)$: [ \begin{align*} h(0.18)&=-3\times(0.18)^{3}+9\times(0.18)^{2}-3\times0.18-4\ &=-3\times0.005832 + 9\times0.0324-0.54 - 4\ &=-0.017496+0.2916-0.54 - 4\ &=-4.265896\approx - 4.27 \end{align*} ]

So the point $(x,h(x))$ where there is a local minimum is $(0.18,-4.27)$

Explanation:

Step1: Find the first - derivative

$h'(x)=-9x^{2}+18x - 3$

Step2: Set the first - derivative equal to zero

$-9x^{2}+18x - 3 = 0$

Step3: Solve the quadratic equation for critical points

$x=\frac{6\pm\sqrt{36 - 12}}{6}=1\pm\frac{\sqrt{6}}{3}$

Step4: Find the second - derivative

$h''(x)=-18x + 18$

Step5: Evaluate the second - derivative at critical points

Determine local minimum by sign of $h''(x)$

Step6: Find the $y$ - value at the local minimum

$h(0.18)\approx - 4.27$