for each problem, find the: x and y intercepts, x - coordinates of the critical points, open intervals wher…

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.\n1) ( y = -\frac{x^{3}}{3}+x^{2} )
Answer
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)