Question

1. use the graph of the function to find the relative maximum and minimum points if they exist. state the maximum and/or minimum points and then use interval notation to state the intervals on which the graph is increasing and the intervals where it is decreasing. round all coordinates to the nearest hundredth. email me or attach a screenshot of the graph for full points

$f(x)=x^{4}-4x^{3}+4x^{2}$

maximum point(s):

minimum point(s):

increasing interval(s):

decreasing interval(s):

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): $(-\infty,0),(1,2)$