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 t 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

For x-intercepts: Set $y=0$ → $-\frac{x^3}{3} + x^2 = 0$ → $x^2(-\frac{x}{3} + 1) = 0$ → $x=0$ or $x=3$.
For y-intercept: Set $x=0$ → $y=0$.

Step2: Find critical points

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

Step3: Increasing/decreasing intervals

Test $y'$:

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

Step4: Relative extrema

At $x=0$: $y'$ changes from - to + → relative minimum ($y=0$).
At $x=2$: $y'$ changes from + to - → relative maximum ($y=-\frac{8}{3} + 4 = \frac{4}{3}$).

Step5: Find inflection points

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

Step6: Concavity intervals

Test $y''$:

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

Answer:

x-intercepts: $0, 3$; y-intercept: $0$
Critical points x-coordinates: $0, 2$
Increasing: $(0, 2)$; Decreasing: $(-\infty, 0) \cup (2, +\infty)$
Relative minimum at $x=0$ (value $0$); Relative maximum at $x=2$ (value $\frac{4}{3}$)
Inflection point x-coordinate: $1$
Concave up: $(-\infty, 1)$; Concave down: $(1, +\infty)$

(Graph sketch: Plot intercepts $(0,0),(3,0)$, extrema $(0,0),(2,\frac{4}{3})$, inflection point $(1,\frac{2}{3})$; curve decreases to $(0,0)$, increases to $(2,\frac{4}{3})$, then decreases; concave up left of $x=1$, concave down right of $x=1$.)