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 \(x\) and \(y\) - intercepts

• \(y\)-intercept:

Set \(x = 0\) in \(y=-\frac{x^{3}}{3}+x^{2}\). Then \(y = 0\).

• \(x\)-intercept:

Set \(y = 0\), so \(0=-\frac{x^{3}}{3}+x^{2}\). Factor out \(x^{2}\): \(x^{2}(1-\frac{x}{3}) = 0\).
Solving \(x^{2}=0\) gives \(x = 0\), and solving \(1-\frac{x}{3}=0\) gives \(x = 3\).

Step2: Find the first - derivative \(y'\) and critical points

• Differentiate \(y=-\frac{x^{3}}{3}+x^{2}\) using the power rule \((x^{n})'=nx^{n - 1}\).

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

Step3: Determine intervals of increase and decrease

• Consider the intervals \((-\infty,0)\), \((0,2)\), and \((2,\infty)\).

Take a test - point in \((-\infty,0)\), say \(x=-1\). Then \(y'=-(-1)^{2}+2(-1)=-1 - 2=-3<0\).
Take a test - point in \((0,2)\), say \(x = 1\). Then \(y'=-1^{2}+2\times1=1>0\).
Take a test - point in \((2,\infty)\), say \(x = 3\). Then \(y'=-3^{2}+2\times3=-9 + 6=-3<0\).
The function is decreasing on \((-\infty,0)\cup(2,\infty)\) and increasing on \((0,2)\).

Step4: Find the second - derivative \(y''\) and inflection points

• Differentiate \(y'=-x^{2}+2x\). Using the power rule, \(y''=-2x + 2\).

Set \(y''=0\), then \(-2x+2=0\). Solving for \(x\) gives \(x = 1\).

Step5: Determine intervals of concavity

• Consider the intervals \((-\infty,1)\) and \((1,\infty)\).

Take a test - point in \((-\infty,1)\), say \(x = 0\). Then \(y''=-2\times0 + 2=2>0\).
Take a test - point in \((1,\infty)\), say \(x = 2\). Then \(y''=-2\times2+2=-2<0\).
The function is concave up on \((-\infty,1)\) and concave down on \((1,\infty)\).

Step6: Find relative minima and maxima

• Use the first - derivative test.

Since \(y'\) changes from negative to positive at \(x = 0\), \(y(0)=0\) is a relative minimum.
Since \(y'\) changes from positive to negative at \(x = 2\), \(y(2)=-\frac{8}{3}+4=\frac{4}{3}\) is a relative maximum.

Answer:

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