Question

the graph of the following function has one relative maximum point and one relative minimum point. find these using the first - derivative test.
f(x)=x^3 + 3x^2 - 9x + 1
find the first derivative of the function.
f(x)=3x^2 + 6x - 9
the relative maximum is (type an ordered pair.)
the relative minimum is (type an ordered pair.)

Explanation:

Step1: Find critical points

Set $f'(x)=3x^{2}+6x - 9 = 0$. Factor out 3: $3(x^{2}+2x - 3)=0$, then $x^{2}+2x - 3=(x + 3)(x - 1)=0$. So $x=-3$ or $x = 1$.

Step2: Use first - derivative test

Choose test points in the intervals $(-\infty,-3)$, $(-3,1)$ and $(1,\infty)$. For $x=-4$ in $(-\infty,-3)$, $f'(-4)=3\times(-4)^{2}+6\times(-4)-9=48-24 - 9 = 15>0$. For $x = 0$ in $(-3,1)$, $f'(0)=3\times0^{2}+6\times0-9=-9<0$. For $x = 2$ in $(1,\infty)$, $f'(2)=3\times2^{2}+6\times2-9=12 + 12-9 = 15>0$.
Since $f(x)$ is increasing on $(-\infty,-3)$ and decreasing on $(-3,1)$, $x=-3$ is a relative maximum. Substitute $x=-3$ into $f(x)$: $f(-3)=(-3)^{3}+3\times(-3)^{2}-9\times(-3)+1=-27 + 27+27 + 1=28$.
Since $f(x)$ is decreasing on $(-3,1)$ and increasing on $(1,\infty)$, $x = 1$ is a relative minimum. Substitute $x = 1$ into $f(x)$: $f(1)=1^{3}+3\times1^{2}-9\times1+1=1 + 3-9 + 1=-4$.

Answer:

The relative maximum is $(-3,28)$
The relative minimum is $(1,-4)$