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

• Use the power rule: \(y'=-x^{2}+2x\).
• 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 \(x=-1\): \(y'=-(-1)^{2}+2(-1)=-1 - 2=-3<0\), 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 \(x = 3\): \(y'=-3^{2}+2\times3=-9 + 6=-3<0\), so the function is decreasing on \((2,\infty)\).

Step4: Use the second - derivative test for relative extrema

• Find the second - derivative \(y''=-2x + 2\).
• At \(x = 0\): \(y''=-2\times0+2 = 2>0\), so \(y(0)=0\) is a relative minimum.
• At \(x = 2\): \(y''=-2\times2+2=-2<0\), so \(y(2)=-\frac{8}{3}+4=\frac{4}{3}\) is a relative maximum.

Step5: Find the second - derivative \(y''\) and inflection points

• Set \(y'' = 0\): \(-2x+2 = 0\), solve for \(x\): \(x = 1\).
• 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 \((1,\infty)\).

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 concave up: \((-\infty,1)\)
• Intervals of concave down: \((1,\infty)\)
• Relative minimum: At \((0,0)\)
• Relative maximum: At \((2,\frac{4}{3})\)