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 intercepts

Set $x=0$: $y=0$. Set $y=0$: $-\frac{x^3}{3}+x^2=0 \Rightarrow x^2(3-x)=0 \Rightarrow x=0,3$.

Step2: Find critical points

First derivative: $y'=-x^2+2x=-x(x-2)$. Set $y'=0$: $x=0,2$.

Step3: Determine increasing/decreasing

Test intervals: $(-\infty,0)$: $y'<0$ (decreasing); $(0,2)$: $y'>0$ (increasing); $(2,\infty)$: $y'<0$ (decreasing).

Step4: Find relative extrema

At $x=0$: decreasing→increasing (min); at $x=2$: increasing→decreasing (max).

Step5: Find inflection points

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

Step6: Determine concavity

Test intervals: $(-\infty,1)$: $y''>0$ (concave up); $(1,\infty)$: $y''<0$ (concave down).

Answer:

• x-intercepts: $0, 3$; y-intercept: $0$
• Critical points x-coordinates: $0, 2$
• Increasing: $(0,2)$; Decreasing: $(-\infty,0)\cup(2,\infty)$
• Inflection point x-coordinate: $1$
• Concave up: $(-\infty,1)$; Concave down: $(1,\infty)$
• Relative min at $x=0$; Relative max at $x=2$
• Graph: Plot points $(0,0)$, $(2,\frac{4}{3})$, $(3,0)$; curve decreasing then increasing then decreasing, concave up then down at $x=1$.