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 (set y=0)

Set $-\frac{x^3}{3} + x^2 = 0$, 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 (set x=0)

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

Step3: Find critical points (solve $y'=0$)

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

Step4: Test increasing/decreasing intervals

Test intervals:

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

Step5: Identify relative min/max

  • At $x=0$: function changes from decreasing to increasing? No, decreasing then increasing → relative minimum? No: decreasing before, increasing after: relative minimum. Wait, correction: $x=0$: left is decreasing, right is increasing → relative min. $x=2$: left increasing, right decreasing → relative max. Calculate values:
  • $x=0$: $y=0$ (relative min)
  • $x=2$: $y = -\frac{8}{3} + 4 = \frac{4}{3}$ (relative max)

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

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

Step7: Test concavity intervals

Test intervals:

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

Answer:

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

For the graph:

  1. Plot intercepts $(0,0)$ and $(3,0)$
  2. Plot relative min $(0,0)$ and relative max $\left(2,\frac{4}{3}\right)$
  3. Plot inflection point $\left(1,\frac{2}{3}\right)$
  4. Draw the curve decreasing on $(-\infty,0)$, increasing on $(0,2)$, decreasing on $(2,\infty)$; concave up on $(-\infty,1)$, concave down on $(1,\infty)$