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 th 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 \((x^{n})^\prime=nx^{n - 1}\), \(y^\prime=-x^{2}+2x\).
Set \(y^\prime = 0\), then \(-x^{2}+2x=0\). Factor out \(-x\): \(-x(x - 2)=0\).
Solving \(-x=0\) gives \(x = 0\), and solving \(x - 2=0\) gives \(x = 2\).

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 \((-\infty,0)\)): \(y^\prime=-(-1)^{2}+2(-1)=-1 - 2=-3<0\).
For \(x = 1\) (in \((0,2)\)): \(y^\prime=-1^{2}+2\times1=1>0\).
For \(x = 3\) (in \((2,\infty)\)): \(y^\prime=-3^{2}+2\times3=-9 + 6=-3<0\).
The function is decreasing on \((-\infty,0)\cup(2,\infty)\) and increasing on \((0,2)\).

Step4: Find the second - derivative and inflection points

• Second - derivative:

Differentiate \(y^\prime=-x^{2}+2x\) with respect to \(x\). \(y^{\prime\prime}=-2x + 2\).
Set \(y^{\prime\prime}=0\), then \(-2x + 2=0\), which gives \(x = 1\).

Step5: Determine concavity

• Test intervals for concavity:

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