8. use the graph of the function to find the relative maximum and minimum points if they exist. state the…

Explanation:

Step1: Find the derivative

Differentiate $f(x)=x^{4}-4x^{3}+4x^{2}$ using the power - rule. The power - rule states that if $y = x^n$, then $y^\prime=nx^{n - 1}$. So, $f^\prime(x)=4x^{3}-12x^{2}+8x$.

Step2: Set the derivative equal to zero

Factor out $4x$ from $f^\prime(x)$: $4x(x^{2}-3x + 2)=0$. Then factor the quadratic: $4x(x - 1)(x - 2)=0$. Solving for $x$, we get $x = 0$, $x=1$, and $x = 2$.

Step3: Use the first - derivative test

Choose test points in the intervals $(-\infty,0)$, $(0,1)$, $(1,2)$, and $(2,\infty)$. For the interval $(-\infty,0)$, let's choose $x=-1$. Then $f^\prime(-1)=4(-1)^{3}-12(-1)^{2}+8(-1)=-4 - 12-8=-24<0$, so the function is decreasing on $(-\infty,0)$. For the interval $(0,1)$, let's choose $x = 0.5$. Then $f^\prime(0.5)=4(0.5)^{3}-12(0.5)^{2}+8(0.5)=4\times0.125-12\times0.25 + 4=0.5-3 + 4=1.5>0$, so the function is increasing on $(0,1)$. For the interval $(1,2)$, let's choose $x = 1.5$. Then $f^\prime(1.5)=4(1.5)^{3}-12(1.5)^{2}+8(1.5)=4\times3.375-12\times2.25+12=13.5-27 + 12=-1.5<0$, so the function is decreasing on $(1,2)$. For the interval $(2,\infty)$, let's choose $x = 3$. Then $f^\prime(3)=4(3)^{3}-12(3)^{2}+8(3)=4\times27-12\times9 + 24=108-108+24=24>0$, so the function is increasing on $(2,\infty)$.

Step4: Find the maximum and minimum points

Evaluate the function at the critical points $x = 0$, $x = 1$, and $x = 2$. $f(0)=0^{4}-4\times0^{3}+4\times0^{2}=0$. $f(1)=1^{4}-4\times1^{3}+4\times1^{2}=1 - 4 + 4=1$. $f(2)=2^{4}-4\times2^{3}+4\times2^{2}=16-32 + 16=0$.

A relative maximum occurs at $x = 1$ and $y = 1$, so the maximum point is $(1,1)$. Relative minimums occur at $x = 0,y = 0$ and $x = 2,y = 0$, so the minimum points are $(0,0)$ and $(2,0)$.

Answer:

Maximum point(s): $(1,1)$ Minimum point(s): $(0,0),(2,0)$ Increasing interval(s): $(0,1),(2,\infty)$ Decreasing interval(s): $(-\infty,0),(1,2)$