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for each problem, find the: x and y intercepts, x - coordinates of the …

Question

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.

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

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 $y =-\frac{x^{3}}{3}+x^{2}$ 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 intervals of increase and decrease

Test intervals $(-\infty,0)$, $(0,2)$ and $(2,\infty)$.
For $x=-1$ in $(-\infty,0)$, $y'=-(-1)^{2}+2(-1)=-3<0$, so the function is decreasing on $(-\infty,0)$.
For $x = 1$ in $(0,2)$, $y'=-1^{2}+2\times1 = 1>0$, so the function is increasing on $(0,2)$.
For $x = 3$ in $(2,\infty)$, $y'=-3^{2}+2\times3=-3<0$, so the function is decreasing on $(2,\infty)$.

Step6: Find the second - derivative

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

Step7: Find inflection points

Set $y'' = 0$. So, $-2x+2 = 0$, which gives $x = 1$.

Step8: Determine concavity

Test intervals $(-\infty,1)$ and $(1,\infty)$.
For $x = 0$ in $(-\infty,1)$, $y''=-2\times0 + 2=2>0$, so the function is concave up on $(-\infty,1)$.
For $x = 2$ in $(1,\infty)$, $y''=-2\times2+2=-2<0$, so the function is concave down on $(1,\infty)$.

Step9: Find relative minima and maxima

Since the function changes from decreasing to increasing at $x = 0$, $y(0)=0$ is a relative minimum.
Since the function changes from increasing to decreasing 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: $x = 0,x = 2$
  • Intervals of increase: $(0,2)$
  • Intervals of decrease: $(-\infty,0)\cup(2,\infty)$
  • Inflection point: $x = 1$
  • Intervals of concavity: Concave up on $(-\infty,1)$, concave down on $(1,\infty)$
  • Relative minimum: At $x = 0,y = 0$
  • Relative maximum: At $x = 2,y=\frac{4}{3}$