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 \(0=-\frac{x^{3}}{3}+x^{2}=x^{2}(1 - \frac{x}{3})\). Solving \(x^{2}(1-\frac{x}{3})=0\) gives \(x = 0\) or \(x = 3\).

Step2: Find the first - derivative and critical points

• First - derivative:

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

• Critical points:

Set \(y^\prime = 0\). Then \(-x(x - 2)=0\), so \(x=0\) or \(x = 2\).

Step3: Determine intervals of increase and decrease

• Test intervals:

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

Step4: Find the second - derivative and inflection points

• Second - derivative:

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

• Inflection points:

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

Step5: Determine intervals of concavity

• Test intervals:

Choose test points in the intervals \((-\infty,1)\) and \((1,\infty)\).
For \(x = 0\) (in \((-\infty,1)\)), \(y^{\prime\prime}=-2\times0+2=2>0\).
For \(x = 2\) (in \((1,\infty)\)), \(y^{\prime\prime}=-2\times2+2=-2<0\).
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 \(y^\prime\) changes from negative to positive at \(x = 0\), \(y(0)=0\) is a relative minimum.
Since \(y^\prime\) 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 \(x = 2\).
• Intervals of increase: \((0,2)\); Intervals of decrease: \((-\infty,0)\cup(2,\infty)\).
• Inflection point: \(x = 1\).
• Intervals of concavity: Concave up on \((-\infty,1)\), concave down on \((1,\infty)\).
• Relative minima: At \(x = 0\), \(y = 0\); Relative maxima: At \(x = 2\), \(y=\frac{4}{3}\).