Question

for each problem, find the: x and y intercepts, x-coordinates of the critical points, open intervals where the function is increasing and decreasing, x-coordinates of the inflection points, open intervals where the function is concave up and concave down, and relative minima and maxima. using this information, sketch the graph of the function. 1) $y = -\frac{x^3}{3} + x^2$

Explanation:

Step1: Find intercepts

Set $x=0$: $y=0$. Set $y=0$: $-\frac{x^3}{3}+x^2=0 \Rightarrow x^2(-\frac{x}{3}+1)=0 \Rightarrow x=0,3$.
Intercepts: $(0,0),(3,0)$.

Step2: Find critical points

First derivative: $y'=-x^2+2x$. Set $y'=0$: $-x(x-2)=0 \Rightarrow x=0,2$.
Critical points at $x=0,2$.

Step3: Determine increasing/decreasing

Test intervals:

• $x<0$: $y'(-1)=-1-2=-3<0$ (decreasing)
• $00$ (increasing)
• $x>2$: $y'(3)=-9+6=-3<0$ (decreasing)

Increasing: $(0,2)$; Decreasing: $(-\infty,0)\cup(2,\infty)$.

Step4: Find relative extrema

$x=0$: decreasing→increasing (min). $x=2$: increasing→decreasing (max).
Relative min at $x=0$; relative max at $x=2$.

Step5: Find inflection points

Second derivative: $y''=-2x+2$. Set $y''=0$: $x=1$.
Inflection point at $x=1$.

Step6: Determine concavity

Test intervals:

• $x<1$: $y''(0)=2>0$ (concave up)
• $x>1$: $y''(2)=-4+2=-2<0$ (concave down)

Concave up: $(-\infty,1)$; Concave down: $(1,\infty)$.

Answer:

Intercepts: $(0,0)$, $(3,0)$
Critical points: $x=0,2$
Increasing: $(0,2)$; Decreasing: $(-\infty,0)\cup(2,\infty)$
Relative min at $x=0$, relative max at $x=2$
Inflection point: $x=1$
Concave up: $(-\infty,1)$; Concave down: $(1,\infty)$

(Graph sketch: Plot intercepts, extrema, inflection point; draw curve decreasing→increasing→decreasing, concave up then down.)