Question

the polynomial function h is defined by h(x)=-3x^3 + 9x^2 - 3x - 4. use the aleks graphing calculator to find all the points (x, h(x)) where there is a local minimum. round to the nearest hundredth. if there is more than one point, enter them using the \and\ button.

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

$h(0.18)\approx - 4.27$

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:

$$
LATEXBLOCK0
$$

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 $c = 1$.

$$
LATEXBLOCK1
$$

$$x_1=1+\frac{\sqrt{6}}{3}\approx1 + 0.82=1.82$$

$$x_2=1-\frac{\sqrt{6}}{3}\approx1-0.82 = 0.18$$

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 $x = 1+\frac{\sqrt{6}}{3}$:

$$
LATEXBLOCK2
$$

For $x = 1-\frac{\sqrt{6}}{3}$:

$$
LATEXBLOCK3
$$

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

Find $h(0.18)$:

$$
LATEXBLOCK4
$$

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