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 y-intercept

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

Step2: Find x-intercepts

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

Step3: Find critical points

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

Step4: Determine increasing/decreasing intervals

Test intervals:

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

Step5: Find relative extrema

By first derivative test:

  • At $x=0$: function changes from decreasing to increasing, so relative minimum
  • At $x=2$: function changes from increasing to decreasing, so relative maximum Calculate extrema values:
  • $y(0) = 0$
  • $y(2) = -\frac{2^3}{3} + 2^2 = -\frac{8}{3} + 4 = \frac{4}{3}$

Step6: Find inflection points

Compute second derivative: $y'' = -2x + 2$ Set $y''=0$: $-2x + 2 = 0$ Solve for $x$: $x=1$ Calculate $y(1) = -\frac{1^3}{3} + 1^2 = \frac{2}{3}$

Step7: Determine concavity intervals

Test intervals:

  • For $(-\infty,1)$: take $x=0$, $y'' = -2(0) + 2 = 2 > 0$, concave up
  • For $(1,+\infty)$: take $x=2$, $y'' = -2(2) + 2 = -2 < 0$, concave down

Answer:

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

(Graph sketch: Plot the intercepts, extrema, inflection point, then connect them following the increasing/decreasing and concavity rules: decreasing concave up from left to $(0,0)$, increasing concave up to $\left(1,\frac{2}{3}\right)$, increasing concave down to $\left(2,\frac{4}{3}\right)$, then decreasing concave down to $(3,0)$ and beyond.)