QuestionSolve the inequality $f(x) \geq 0$ for $f(x)=(x+1)(x-3)^{2}$ using its graph. Provide the solution in interval notation.

Studdy Solution

STEP 1

Assumptions1. The function is $f(x)=(x+1)(x-3)^{}$ . We need to find the solution set for $f(x) \geq0$
3. The solution will be in interval notation

STEP 2

The function $f(x)$ is a product of two factors, $(x+1)$ and $(x-)^{2}$ . The function will be zero when either of these factors is zero. Therefore, we can find the roots of the function by setting each factor equal to zero.
$x+1=0$ $x-=0$

STEP 3

olving the above equations will give us the roots of the function.
$x=-1$ $x=3$

STEP 4

The roots divide the number line into three intervals $(-\infty, -1)$ , $(-1,3)$ , and $(3, \infty)$ . We will test each interval to see where the function is greater than or equal to zero.

STEP 5

Choose a test point in the interval $(-\infty, -1)$ , say $x=-2$ . Substitute $x=-2$ into the function $f(x)$ .
$f(-2)=(-2+1)((-2)-3)^{2}$

STEP 6

Calculate the value of $f(-2)$ .
$f(-2)=(-1)(-5)^{2} = -25$ Since $f(-2) <0$ , the function is negative in the interval $(-\infty, -1)$ .

STEP 7

Choose a test point in the interval $(-1,3)$ , say $x=1$ . Substitute $x=1$ into the function $f(x)$ .
$f(1)=(1+1)((1)-3)^{2}$

STEP 8

Calculate the value of $f(1)$ .
$f(1)=(2)(-2)^{2} =8$ Since $f(1) >0$ , the function is positive in the interval $(-1,3)$ .

STEP 9

Choose a test point in the interval $(3, \infty)$ , say $x=4$ . Substitute $x=4$ into the function $f(x)$ .
$f(4)=(4+)((4)-3)^{2}$

STEP 10

Calculate the value of $f(4)$ .
$f(4)=(5)()^{2} =5$ Since $f(4) >0$ , the function is positive in the interval $(3, \infty)$ .

STEP 11

The solution set for $f(x) \geq0$ is the union of the intervals where the function is positive and the points where the function is zero. Therefore, the solution set is $[-,3] \cup [3, \infty)$ .
However, since the point $x=3$ is repeated in both intervals, we can simplify the solution set to $[-, \infty)$ .
The solution set for $f(x) \geq0$ is $[-, \infty)$ .

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