Question

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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 \(y'\) and critical points

• Differentiate \(y=-\frac{x^{3}}{3}+x^{2}\) using the power rule \((x^{n})'=nx^{n - 1}\).

\(y'=-x^{2}+2x\).
Set \(y'=0\): \(-x^{2}+2x=0\). Factor out \(-x\): \(-x(x - 2)=0\).
Critical points at \(x = 0\) and \(x = 2\).

Step3: Determine intervals of increase and decrease

• Use test intervals.

For \(y'=-x^{2}+2x\), the critical points divide the real line into intervals \((-\infty,0)\), \((0,2)\), and \((2,\infty)\).

• Test \(x=-1\) in \(y'\): \(y'(-1)=-(-1)^{2}+2(-1)=-3<0\), so \(y\) is decreasing on \((-\infty,0)\).
• Test \(x = 1\) in \(y'\): \(y'(1)=-1^{2}+2(1)=1>0\), so \(y\) is increasing on \((0,2)\).
• Test \(x = 3\) in \(y'\): \(y'(3)=-3^{2}+2(3)=-3<0\), so \(y\) is decreasing on \((2,\infty)\).

Step4: Find the second - derivative \(y''\) and inflection points

• Differentiate \(y'=-x^{2}+2x\). \(y''=-2x + 2\).

Set \(y''=0\): \(-2x + 2=0\), then \(x = 1\).

Step5: Determine intervals of concavity

• Test intervals \((-\infty,1)\) and \((1,\infty)\).
• Test \(x = 0\) in \(y''\): \(y''(0)=-2(0)+2=2>0\), so \(y\) is concave up on \((-\infty,1)\).
• Test \(x = 2\) in \(y''\): \(y''(2)=-2(2)+2=-2<0\), so \(y\) is concave down on \((1,\infty)\).

Step6: Find relative minima and maxima

• Use the first - derivative test.

Since \(y'\) changes from negative to positive at \(x = 0\), \(y(0)=0\) is a relative minimum.
Since \(y'\) 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 concave up: \((-\infty,1)\)
• Intervals of concave down: \((1,\infty)\)
• Relative minimum: At \(x = 0\), \(y = 0\)
• Relative maximum: At \(x = 2\), \(y=\frac{4}{3}\)