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

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

Step4: Determine increasing/decreasing

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 min)
• $x=2$: increasing→decreasing (relative max)

Step6: Find inflection points

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

Step7: Determine concavity

Test intervals:

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

Answer:

x-intercepts: 0, 3
y-intercept: 0
Critical points x-coordinates: 0, 2
Increasing interval: (0, 2)
Decreasing intervals: (-∞, 0), (2, ∞)
Relative minimum at x=0, relative maximum at x=2
Inflection point x-coordinate: 1
Concave up interval: (-∞, 1)
Concave down interval: (1, ∞)
(Graph: Plot intercepts (0,0),(3,0); relative min (0,0), relative max (2, 4/3); inflection point (1, 2/3); curve decreasing then increasing then decreasing, concave up then down.)