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 and critical points

• First - derivative:

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

• Intervals of increase and decrease:

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 \((2,\infty)\).
• Relative extrema:

Using the first - derivative test:
At \(x = 0\), since the function changes from decreasing (\(x\lt0\)) to increasing (\(0\lt x\lt2\)) is incorrect (it should be from decreasing to increasing at \(x = 0\) is wrong, actually at \(x = 0\) \(y'=0\) and the sign of \(y'\) changes from negative (\(x\lt0\)) to positive (\(0\lt x\lt2\)), but since \(y''(x)=-2x + 2\), \(y''(0)=2\gt0\), so \(y(0)=0\) is a relative minimum.
At \(x = 2\), since the function changes from increasing (\(0\lt x\lt2\)) to decreasing (\(x\gt2\)), and \(y''(2)=-2\times2 + 2=-2\lt0\), \(y(2)=-\frac{8}{3}+4=\frac{4}{3}\) is a relative maximum.

Step3: Find the second - derivative and inflection points

• Second - derivative:

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

• Intervals of concavity:
• For \(x\lt1\), let \(x = 0\). Then \(y''(0)=2\gt0\), so the function is concave up on \((-\infty,1)\).
• For \(x\gt1\), let \(x = 2\). Then \(y''(2)=-2\lt0\), 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 \(x = 2\)
• Intervals of increase: \((0,2)\)
• Intervals of decrease: \((-\infty,0)\cup(2,\infty)\)
• Relative minimum: At \(x = 0\), \(y(0)=0\)
• Relative maximum: At \(x = 2\), \(y(2)=\frac{4}{3}\)
• Inflection point: \(x = 1\)
• Intervals of concave up: \((-\infty,1)\)
• Intervals of concave down: \((1,\infty)\)

To sketch the graph:

• Plot the intercepts \((0,0)\) and \((3,0)\).
• Plot the relative minimum \((0,0)\) and relative maximum \((2,\frac{4}{3})\).
• Mark the inflection point \((1,y(1))\) where \(y(1)=-\frac{1}{3}+1=\frac{2}{3}\).
• Use the intervals of increase/decrease and concavity to draw the smooth curve of \(y =-\frac{x^{3}}{3}+x^{2}\).