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

Step2: Find critical points

Compute $y'$: $y'=-x^2+2x$.
Set $y'=0$: $-x(x-2)=0 \Rightarrow x=0$ or $x=2$.
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

At $x=0$: decreasing→increasing (min).
At $x=2$: increasing→decreasing (max).
Relative min: $(0,0)$; Relative max: $(2,\frac{4}{3})$.

Step5: Find inflection points

Compute $y''$: $y''=-2x+2$.
Set $y''=0$: $-2x+2=0 \Rightarrow x=1$.
Inflection point at $x=1$.

Step6: Determine concavity

Test intervals:

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

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

Answer:

Intercepts: $(0,0),(3,0)$
Critical points at $x=0,2$
Increasing: $(0,2)$; Decreasing: $(-\infty,0)\cup(2,\infty)$
Relative min: $(0,0)$; Relative max: $(2,\frac{4}{3})$
Inflection point at $x=1$
Concave up: $(-\infty,1)$; Concave down: $(1,\infty)$
(Graph: Passes through (0,0),(3,0); has min at (0,0), max at (2,4/3); inflection at (1,2/3); concave up left of x=1, concave down right of x=1)