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

Using the power rule \((x^{n})'=nx^{n - 1}\), \(y'=-x^{2}+2x=-x(x - 2)\)

• Critical points: Set \(y'=0\). So \(-x(x - 2)=0\), which gives \(x = 0\) and \(x = 2\)
• Increasing/decreasing intervals:
• Test intervals: \((-\infty,0)\), \((0,2)\), \((2,\infty)\)
• 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 \((2,\infty)\)
• Relative minima and maxima:
• Using the first - derivative test: At \(x = 0\), since the function changes from decreasing (\(y'<0\) for \(x<0\)) to increasing (\(y'>0\) for \(00\) for \(00\) for \(02\)), \(y(2)=-\frac{8}{3}+4=\frac{4}{3}\) is a relative maximum.

Step3: Find the second - derivative \(y''\)

Differentiate \(y'=-x^{2}+2x\) with respect to \(x\). Using the power rule, \(y''=-2x + 2=-2(x - 1)\)

• Inflection point: Set \(y'' = 0\). Then \(-2(x - 1)=0\), so \(x = 1\). When \(x = 1\), \(y=-\frac{1}{3}+1=\frac{2}{3}\)
• 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 \((1,\infty)\)

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: At \(x = 0\), \(y = 0\)
• Relative maximum: At \(x = 2\), \(y=\frac{4}{3}\)

To sketch the graph:

• Plot the intercepts \((0,0)\) and \((3,0)\)
• Mark the critical points \((0,0)\) (relative minimum) and \((2,\frac{4}{3})\) (relative maximum)
• Mark the inflection point \((1,\frac{2}{3})\)
• Use the information about increasing/decreasing (function decreases on \((-\infty,0)\), increases on \((0,2)\), decreases on \((2,\infty)\)) and concavity (concave up on \((-\infty,1)\), concave down on \((1,\infty)\)) to draw a smooth curve.