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 \(y=-\frac{0^{3}}{3}+0^{2}=0\).

• \(x\)-intercept:

Set \(y = 0\), so \(0=-\frac{x^{3}}{3}+x^{2}\). Factor out \(x^{2}\): \(x^{2}(1-\frac{x}{3}) = 0\).
Using the zero - product property \(x^{2}=0\) gives \(x = 0\), and \(1-\frac{x}{3}=0\) gives \(x = 3\).

Step2: Find the first derivative and critical points

• First derivative:

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

• Critical points:

Set \(y^\prime = 0\), so \(-x^{2}+2x=0\). Factor out \(-x\): \(-x(x - 2)=0\).
By the zero - product property, \(x = 0\) or \(x = 2\).

Step3: Determine intervals of increase and decrease

• Test intervals:

The critical points \(x = 0\) and \(x = 2\) divide the real line into intervals \((-\infty,0)\), \((0,2)\), and \((2,\infty)\).

• For \(x\in(-\infty,0)\), let \(x=-1\). Then \(y^\prime=-(-1)^{2}+2(-1)=-1 - 2=-3<0\), so the function is decreasing on \((-\infty,0)\).
• For \(x\in(0,2)\), let \(x = 1\). Then \(y^\prime=-1^{2}+2\times1=1>0\), so the function is increasing on \((0,2)\).
• For \(x\in(2,\infty)\), let \(x = 3\). Then \(y^\prime=-3^{2}+2\times3=-9 + 6=-3<0\), so the function is decreasing on \((2,\infty)\).

Step4: Find the second derivative and inflection points

• Second derivative:

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

• Inflection points:

Set \(y^{\prime\prime}=0\), so \(-2x + 2=0\). Solving for \(x\) gives \(x = 1\).

Step5: Determine intervals of concavity

• Test intervals:

The inflection point \(x = 1\) divides the real line into intervals \((-\infty,1)\) and \((1,\infty)\).

• For \(x\in(-\infty,1)\), let \(x = 0\). Then \(y^{\prime\prime}=-2\times0+2=2>0\), so the function is concave up on \((-\infty,1)\).
• For \(x\in(1,\infty)\), let \(x = 2\). Then \(y^{\prime\prime}=-2\times2+2=-2<0\), so the function is concave down on \((1,\infty)\).

Step6: Find relative minima and maxima

• Use the first - derivative test:

Since \(y^\prime\) changes sign from negative (\(x\in(-\infty,0)\)) to positive (\(x\in(0,2)\)) at \(x = 0\), \(y(0)=0\) is a relative minimum.
Since \(y^\prime\) changes sign from positive (\(x\in(0,2)\)) to negative (\(x\in(2,\infty)\)) 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: \(x = 0\) 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: At \(x = 0\), \(y = 0\)
• Relative maximum: At \(x = 2\), \(y=\frac{4}{3}\)