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

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.\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: Find critical points
First derivative: $y' = -x^2 + 2x$
Set $y'=0$: $-x(x-2)=0 \Rightarrow x=0$ or $x=2$
Step4: 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)
Step5: Find relative extrema
- $x=0$: decreasing→increasing (minima)
- $x=2$: increasing→decreasing (maxima)
Calculate values: $y(0)=0$, $y(2)=-\frac{8}{3}+4=\frac{4}{3}$
Step6: Find inflection points
Second derivative: $y''=-2x + 2$
Set $y''=0$: $x=1$
Step7: Determine concavity
Test intervals:
- $(-\infty,1)$: $y''(0)=2>0$ (concave up)
- $(1,\infty)$: $y''(2)=-2<0$ (concave down)
Step8: Sketch the graph
Plot intercepts $(0,0),(3,0)$, extrema $(0,0),(2,\frac{4}{3})$, inflection point $(1,\frac{2}{3})$; use intervals for shape.
Answer:
x-intercepts: $0,3$; y-intercept: $0$; critical points at $x=0,2$; increasing on $(0,2)$, decreasing on $(-\infty,0)\cup(2,\infty)$; relative minima at $(0,0)$, relative maxima at $(2,\frac{4}{3})$; inflection point at $x=1$; concave up on $(-\infty,1)$, concave down on $(1,\infty)$; graph as described.