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

• The first - derivative \(y^\prime\) 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 \(x\): \(x(x - 2)=0\). The critical points are \(x = 0\) and \(x = 2\).

• Intervals of increase and decrease:

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

• Relative minima and maxima:

Using the first - derivative test, 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.

Step3: Find the second - derivative and inflection points

• The second - derivative \(y^{\prime\prime}=-2x + 2\). Set \(y^{\prime\prime}=0\), then \(2x=2\), \(x = 1\).
• Intervals of concavity:

Test intervals \((-\infty,1)\) and \((1,\infty)\).
For \(x = 0\), \(y^{\prime\prime}=2>0\), so the function is concave up on \((-\infty,1)\).
For \(x = 2\), \(y^{\prime\prime}=-2<0\), 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 concavity: Concave up on \((-\infty,1)\), concave down on \((1,\infty)\)
• Relative minimum: \((0,0)\)
• Relative maximum: \((2,\frac{4}{3})\)

To sketch the graph:

• Plot the intercepts \((0,0)\) and \((3,0)\).
• Plot the relative minimum \((0,0)\) and relative maximum \((2,\frac{4}{3})\).
• Note the concavity changes at \(x = 1\). Connect the points with a smooth curve, decreasing on \((-\infty,0)\), increasing on \((0,2)\), and decreasing on \((2,\infty)\), concave up on \((-\infty,1)\) and concave down on \((1,\infty)\).