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 $y = 0$, so $-\frac{x^{3}}{3}+x^{2}=0$. Factor out $x^{2}$: $x^{2}(1 - \frac{x}{3})=0$. Then $x = 0$ or $x = 3$.

Step2: Find y - intercepts

Set $x = 0$, then $y=-\frac{0^{3}}{3}+0^{2}=0$.

Step3: Find the first - derivative

Differentiate $y =-\frac{x^{3}}{3}+x^{2}$ with respect to $x$. Using the power rule, $y'=-x^{2}+2x$.

Step4: Find critical points

Set $y' = 0$, so $-x^{2}+2x=0$. Factor out $-x$: $-x(x - 2)=0$. The critical points are $x = 0$ and $x = 2$.

Step5: Determine increasing and decreasing intervals

Test intervals: For $x\lt0$, let $x=-1$, then $y'=-(-1)^{2}+2(-1)=-1 - 2=-3\lt0$, so the function is decreasing on $(-\infty,0)$. For $0\lt x\lt2$, let $x = 1$, then $y'=-1^{2}+2\times1=1\gt0$, so the function is increasing on $(0,2)$. For $x\gt2$, let $x = 3$, then $y'=-3^{2}+2\times3=-9 + 6=-3\lt0$, so the function is decreasing on $(2,\infty)$.

Step6: Find the second - derivative

Differentiate $y'=-x^{2}+2x$ with respect to $x$. $y''=-2x + 2$.

Step7: Find inflection points

Set $y'' = 0$, so $-2x+2 = 0$. Solving for $x$ gives $x = 1$.

Step8: Determine concavity

For $x\lt1$, let $x = 0$, then $y''=-2\times0 + 2=2\gt0$, so the function is concave up on $(-\infty,1)$. For $x\gt1$, let $x = 2$, then $y''=-2\times2+2=-2\lt0$, so the function is concave down on $(1,\infty)$.

Step9: Find relative extrema

Since the function changes from decreasing ($x\lt0$) to increasing ($0\lt x\lt2$) at $x = 0$, $y(0)=0$ is a relative minimum. Since the function changes from increasing ($0\lt x\lt2$) to decreasing ($x\gt2$) at $x = 2$, $y(2)=-\frac{2^{3}}{3}+2^{2}=-\frac{8}{3}+4=\frac{4}{3}$ is a relative maximum.

Answer:

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