Question

questions 20 - 21, find the intercepts and relative extrema for the graph of each function.

1. $f(x)=\frac{1}{4}(x + 2)(x - 1)^2$ (1 point)

Explanation:

Step1: Find the x - intercepts

Set $f(x)=0$.
$\frac{1}{4}(x + 2)(x - 1)^2=0$.
Using the zero - product property, if $ab = 0$, then $a = 0$ or $b = 0$. So $x+2=0$ gives $x=-2$ and $(x - 1)^2=0$ gives $x = 1$. The x - intercepts are $x=-2$ and $x = 1$.

Step2: Find the y - intercept

Set $x = 0$.
$f(0)=\frac{1}{4}(0 + 2)(0 - 1)^2=\frac{1}{4}\times2\times1=\frac{1}{2}$. The y - intercept is $y=\frac{1}{2}$.

Step3: Find the derivative of $f(x)$

First, expand $f(x)=\frac{1}{4}(x + 2)(x^2-2x + 1)=\frac{1}{4}(x^3-2x^2+x+2x^2-4x + 2)=\frac{1}{4}(x^3-3x + 2)$.
Then, using the power rule $(x^n)'=nx^{n - 1}$, $f'(x)=\frac{1}{4}(3x^2-3)=\frac{3}{4}(x^2 - 1)=\frac{3}{4}(x + 1)(x - 1)$.

Step4: Find the critical points

Set $f'(x)=0$.
$\frac{3}{4}(x + 1)(x - 1)=0$.
Using the zero - product property, $x=-1$ or $x = 1$.

Step5: Determine the relative extrema

Use the first - derivative test.
Choose test points in the intervals $(-\infty,-1)$, $(-1,1)$ and $(1,\infty)$.
For the interval $(-\infty,-1)$, let $x=-2$. Then $f'(-2)=\frac{3}{4}((-2)^2-1)=\frac{3}{4}(4 - 1)=\frac{9}{4}>0$, so $f(x)$ is increasing on $(-\infty,-1)$.
For the interval $(-1,1)$, let $x = 0$. Then $f'(0)=\frac{3}{4}(0^2-1)=-\frac{3}{4}<0$, so $f(x)$ is decreasing on $(-1,1)$.
For the interval $(1,\infty)$, let $x = 2$. Then $f'(2)=\frac{3}{4}(2^2-1)=\frac{9}{4}>0$, so $f(x)$ is increasing on $(1,\infty)$.
At $x=-1$, $f(-1)=\frac{1}{4}(-1 + 2)(-1 - 1)^2=\frac{1}{4}\times1\times4 = 1$, so there is a relative maximum at $(-1,1)$.
At $x = 1$, $f(1)=\frac{1}{4}(1 + 2)(1 - 1)^2=0$, so there is a relative minimum at $(1,0)$.

Answer:

x - intercepts: $x=-2, x = 1$; y - intercept: $y=\frac{1}{2}$; relative maximum: $(-1,1)$; relative minimum: $(1,0)$