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 x-intercepts

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

Step2: Find y-intercept

Set $x=0$: $y=-\frac{0^3}{3} + 0^2 = 0$

Step3: Find critical points

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

Step4: Determine increasing/decreasing intervals

Test intervals:

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

Step5: Find relative extrema

• $x=0$: decreasing→increasing (relative minimum)
• $x=2$: increasing→decreasing (relative maximum)

Calculate values: $y(0)=0$, $y(2)=-\frac{8}{3}+4=\frac{4}{3}$

Step6: Find inflection points

Second derivative: $y''=-2x + 2$
Set $y''=0$: $-2x + 2=0 \Rightarrow x=1$
$y(1)=-\frac{1}{3}+1=\frac{2}{3}$

Step7: Determine concavity intervals

Test intervals:

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

Answer:

x-intercepts: $x=0, 3$; y-intercept: $y=0$
Critical points: $x=0, 2$
Increasing: $(0,2)$; Decreasing: $(-\infty,0)\cup(2,\infty)$
Relative minimum: $(0,0)$; Relative maximum: $(2,\frac{4}{3})$
Inflection point: $(1,\frac{2}{3})$
Concave up: $(-\infty,1)$; Concave down: $(1,\infty)$
(Graph: Plot intercepts (0,0),(3,0), extrema (0,0),(2,4/3), inflection point (1,2/3); connect with curve decreasing then increasing then decreasing, concave up then down.)