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

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

Step2: Find critical points

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

Step3: Determine increasing/decreasing

Test intervals:

• $(-\infty,0)$: $y'<0$ (decreasing).
• $(0,2)$: $y'>0$ (increasing).
• $(2,\infty)$: $y'<0$ (decreasing).

Step4: Find relative extrema

At $x=0$: $y=0$ (relative min).
At $x=2$: $y=-\frac{8}{3} + 4 = \frac{4}{3}$ (relative max).

Step5: Find inflection points

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

Step6: Determine concavity

Test intervals:

• $(-\infty,1)$: $y''>0$ (concave up).
• $(1,\infty)$: $y''<0$ (concave down).

Answer:

x-intercepts: $0, 3$; y-intercept: $0$
Critical points: $x=0, 2$
Increasing: $(0,2)$; Decreasing: $(-\infty,0)\cup(2,\infty)$
Relative min: $(0,0)$; Relative max: $(2,\frac{4}{3})$
Inflection point: $(1,\frac{2}{3})$
Concave up: $(-\infty,1)$; Concave down: $(1,\infty)$

(Graph sketch: Plot intercepts (0,0),(3,0), extrema (0,0),(2,4/3), inflection point (1,2/3); connect with curve decreasing→increasing→decreasing, concave up→concave down at x=1.)