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 \(x^{2}\): \(x^{2}(1-\frac{x}{3}) = 0\).
Solutions are \(x = 0\) and \(x = 3\).

Step2: Find the first - derivative and critical points

The first - derivative \(y^\prime=-x^{2}+2x\).
Set \(y^\prime = 0\), 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

• Test intervals:

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

• Relative extrema:

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 \(x = 2\), \(y(2)=-\frac{8}{3}+4=\frac{4}{3}\) is a relative maximum.

Step4: Find the second - derivative and inflection points

The second - derivative \(y^{\prime\prime}=-2x + 2\).
Set \(y^{\prime\prime}=0\), then \(-2x + 2=0\), which gives \(x = 1\).

Step5: Determine intervals of concavity

• Test intervals:

Take the intervals \((-\infty,1)\) and \((1,\infty)\).
For \(x = 0\) (in \((-\infty,1)\)), \(y^{\prime\prime}=-2\times0+2 = 2>0\). So the function is concave up on \((-\infty,1)\).
For \(x = 2\) (in \((1,\infty)\)), \(y^{\prime\prime}=-2\times2+2=-2<0\). So the function is concave down on \((1,\infty)\).

Answer:

• \(x\)-intercepts: \(x = 0\) and \(x = 3\)
• \(y\)-intercept: \(y = 0\)
• Critical points (\(x\)-coordinates): \(x = 0\) and \(x = 2\)
• Intervals of increase: \((0,2)\)
• Intervals of decrease: \((-\infty,0)\cup(2,\infty)\)
• Inflection point (\(x\)-coordinate): \(x = 1\)
• Intervals of concave up: \((-\infty,1)\)
• Intervals of concave down: \((1,\infty)\)
• Relative minimum: \((0,0)\)
• Relative maximum: \((2,\frac{4}{3})\)