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

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

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

Answer

Explanation:

Step1: Find x-intercepts (y=0)

Set $y=0$: $$-\frac{x^3}{3} + x^2 = 0$$ Factor: $$x^2\left(-\frac{x}{3} + 1\right) = 0$$ Solve for $x$: $x^2=0 \implies x=0$; $-\frac{x}{3}+1=0 \implies x=3$

Step2: Find y-intercept (x=0)

Substitute $x=0$: $$y=-\frac{0^3}{3} + 0^2 = 0$$

Step3: Find critical points (y'=0)

Compute first derivative: $$y' = -x^2 + 2x$$ Set $y'=0$: $$-x^2 + 2x = 0 \implies x(-x+2)=0$$ Solve for $x$: $x=0$; $x=2$

Step4: Test increasing/decreasing intervals

Use critical points to split domain:

  • For $(-\infty,0)$: pick $x=-1$, $y'=-(-1)^2+2(-1)=-3<0$ (decreasing)
  • For $(0,2)$: pick $x=1$, $y'=-(1)^2+2(1)=1>0$ (increasing)
  • For $(2,\infty)$: pick $x=3$, $y'=-(3)^2+2(3)=-3<0$ (decreasing)

Step5: Find relative extrema

Use First Derivative Test:

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

Step6: Find inflection points (y''=0)

Compute second derivative: $$y'' = -2x + 2$$ Set $y''=0$: $$-2x+2=0 \implies x=1$$

Step7: Test concavity intervals

Use inflection point to split domain:

  • For $(-\infty,1)$: pick $x=0$, $y''=2>0$ (concave up)
  • For $(1,\infty)$: pick $x=2$, $y''=-2<0$ (concave down) Compute $y(1)=-\frac{1}{3}+1=\frac{2}{3}$

Answer:

  1. Intercepts:
    • x-intercepts: $(0,0)$ and $(3,0)$
    • y-intercept: $(0,0)$
  2. Critical points (x-coordinates): $x=0$, $x=2$
  3. Increasing/decreasing intervals:
    • Decreasing: $(-\infty, 0) \cup (2, \infty)$
    • Increasing: $(0, 2)$
  4. Inflection point (x-coordinate): $x=1$ (point: $(1, \frac{2}{3})$)
  5. Concavity intervals:
    • Concave up: $(-\infty, 1)$
    • Concave down: $(1, \infty)$
  6. Relative extrema:
    • Relative minimum: $(0, 0)$
    • Relative maximum: $(2, \frac{4}{3})$

(Graph sketch guidance: Plot intercepts, extrema, inflection point; draw decreasing curve from left to $(0,0)$, increasing to $(2,\frac{4}{3})$, then decreasing; curve is concave up left of $x=1$, concave down right of $x=1$)