Question

for each problem, find the: x and y intercepts, x - coordinates of the critical points, open intervals whe the function is increasing and decreasing, x - coordinates of the inflection points, open intervals where 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

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

Step4: Intervals of increase/decrease

Test $y'$:

• $x<0$: $y' < 0$ (decreasing)
• $0 0$ (increasing)
• $x>2$: $y' < 0$ (decreasing)

Step5: Relative extrema

• $x=0$: $y'$ changes from - to + → relative minimum
• $x=2$: $y'$ changes from + to - → relative maximum

Compute values:

• $y(0)=0$ (min), $y(2)=-\frac{8}{3} + 4 = \frac{4}{3}$ (max)

Step6: Find inflection points

Compute $y''$: $y'' = -2x + 2$
Set $y''=0$: $-2x + 2 = 0 \Rightarrow x=1$

Step7: Concavity intervals

Test $y''$:

• $x<1$: $y''>0$ (concave up)
• $x>1$: $y''<0$ (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 minimum: $(0, 0)$
Relative maximum: $(2, \frac{4}{3})$
Inflection point (x): $1$
Concave up: $(-\infty, 1)$
Concave down: $(1, +\infty)$

(Sketch: Plot intercepts (0,0),(3,0); extrema (0,0),(2,4/3); inflection at x=1 (y=-1/3 +1=2/3); curve decreasing left of 0, increasing between 0-2, decreasing right of 2; concave up left of 1, concave down right of 1.)