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 \(x = 0\). Then \(y=-\frac{0^{3}}{3}+0^{2}=0\).
• \(x\) - intercept: Set \(y = 0\). So \(0=-\frac{x^{3}}{3}+x^{2}\), factor out \(x^{2}\): \(x^{2}(1 - \frac{x}{3})=0\). Solving \(x^{2}=0\) gives \(x = 0\), and solving \(1-\frac{x}{3}=0\) gives \(x = 3\).

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 \(-x\): \(-x(x - 2)=0\). The critical points are \(x = 0\) and \(x = 2\).

Step3: Determine intervals of increase and decrease

• Consider the intervals \((-\infty,0)\), \((0,2)\), and \((2,\infty)\).
• Test a value in \((-\infty,0)\), say \(x=-1\): \(y'=-(-1)^{2}+2(-1)=-3<0\), so the function is decreasing on \((-\infty,0)\).
• Test a value in \((0,2)\), say \(x = 1\): \(y'=-1^{2}+2\times1 = 1>0\), so the function is increasing on \((0,2)\).
• Test a value in \((2,\infty)\), say \(x = 3\): \(y'=-3^{2}+2\times3=-3<0\), so the function is decreasing on \((2,\infty)\).

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

• Differentiate \(y'=-x^{2}+2x\). \(y''=-2x + 2\).
• Set \(y''=0\): \(-2x+2=0\), solve for \(x\): \(x = 1\).

Step5: Determine intervals of concavity

• Consider the intervals \((-\infty,1)\) and \((1,\infty)\).
• Test a value in \((-\infty,1)\), say \(x = 0\): \(y''=-2\times0+2=2>0\), so the function is concave up on \((-\infty,1)\).
• Test a value in \((1,\infty)\), say \(x = 2\): \(y''=-2\times2+2=-2<0\), so the function is concave down on \((1,\infty)\).

Step6: Find relative minima and maxima

• Use the first - derivative test.
• At \(x = 0\): Since the function changes from decreasing \((x<0)\) to increasing \((0<x<2)\) is incorrect (it should be from decreasing on \((-\infty,0)\) to increasing on \((0,2)\)), but \(y(0)=0\).
• At \(x = 2\): Since the function changes from increasing \((0<x<2)\) to decreasing \((x>2)\), \(y(2)=-\frac{8}{3}+4=\frac{4}{3}\). So \(x = 2\) is a relative maximum and \(x = 0\) is neither (because the sign of \(y'\) changes from negative to positive around \(x = 0\) but \(y'\) is \(0\) at \(x = 0\) and the function has a horizontal tangent, but the second - derivative test: \(y''(0)=2>0\) (but since \(y'(0) = 0\) and \(y''(0)>0\), it's a local minimum. Wait, no, first - derivative test: \(y'(-1)=-3<0\), \(y'(1)=1>0\), so \(x = 0\) is a local minimum. \(y(0)=0\)).

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 minima: At \(x = 0\), \(y = 0\)
• Relative maxima: At \(x = 2\), \(y=\frac{4}{3}\)