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 $x=0$: $y=0$

Step3: Compute first derivative

$y' = -x^2 + 2x$

Step4: Find critical points

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

Step5: Determine increasing/decreasing

Test intervals:

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

Step6: Find relative extrema

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

Step7: Compute second derivative

$y'' = -2x + 2$

Step8: Find inflection points

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

Step9: Determine concavity

Test intervals:

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

Answer:

x-intercepts: 0, 3; y-intercept: 0
Critical points: 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: $(1,\infty)$
(Sketch: 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)