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 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 $x=0$: $y=0 + 0 = 0$

Step3: Find critical points

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

Step4: Find increasing/decreasing intervals

Test $y'$ sign:

• $x<0$: $y'<0$ (decreasing)
• $00$ (increasing)
• $x>2$: $y'<0$ (decreasing)

Step5: Find inflection points

Compute $y''$: $y'' = -2x + 2$
Set $y''=0$: $-2(x - 1)=0 \Rightarrow x=1$

Step6: Find concavity intervals

Test $y''$ sign:

• $x<1$: $y''>0$ (concave up)
• $x>1$: $y''<0$ (concave down)

Step7: Find relative extrema

• $x=0$: $y'$ changes from - to + → relative min
• $x=2$: $y'$ changes from + to - → relative max

Compute values:

• Relative min at $(0,0)$
• Relative max at $(2, -\frac{8}{3} + 4 = \frac{4}{3})$

Answer:

x-intercepts: $x=0, 3$
y-intercept: $y=0$
Critical points x-coordinates: $0, 2$
Increasing interval: $(0,2)$
Decreasing intervals: $(-\infty,0), (2,\infty)$
Inflection point x-coordinate: $1$
Concave up interval: $(-\infty,1)$
Concave down interval: $(1,\infty)$
Relative min: $(0,0)$
Relative max: $(2,\frac{4}{3})$

(Graph sketch: Plot intercepts (0,0),(3,0), extrema (0,0),(2,4/3), inflection point (1, -1/3 +1=2/3); curve decreases from left to (0,0), increases to (2,4/3), decreases after; concave up before x=1, concave down after.)