the graph of the following function has one relative maximum point and one relative minimum point. find…

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

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

Answer

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 $(1,-4)$