Question

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

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

Step6: Identify relative extrema

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

Step7: Compute second derivative

$y''=-2x + 2$

Step8: Find inflection points

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

Step9: Determine concavity intervals

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: $(1,\infty)$
(Graph: Plot intercepts, extrema, inflection point; curve decreasing then increasing then decreasing, concave up then down)