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 \(y'\)

Differentiate \(y=-\frac{x^{3}}{3}+x^{2}\) using the power rule \(y'=-x^{2}+2x\).
Set \(y' = 0\) for critical points: \(-x^{2}+2x=0\), factor out \(-x\): \(-x(x - 2)=0\). So \(x = 0\) and \(x = 2\) are critical points.

• Test intervals for increasing/decreasing:
• Choose a test point in \((-\infty,0)\), say \(x=-1\). Then \(y'=-(-1)^{2}+2(-1)=-1 - 2=-3<0\), so the function is decreasing on \((-\infty,0)\).
• Choose a test point in \((0,2)\), say \(x = 1\). Then \(y'=-1^{2}+2\times1=1>0\), so the function is increasing on \((0,2)\).
• Choose a test point in \((2,\infty)\), say \(x = 3\). Then \(y'=-3^{2}+2\times3=-9 + 6=-3<0\), so the function is decreasing on \((2,\infty)\).
• Relative minima and maxima:

Since the function changes from decreasing \((-\infty,0)\) to increasing \((0,2)\), at \(x = 0\) there is a relative minimum. \(y(0)=0\).
Since the function changes from increasing \((0,2)\) to decreasing \((2,\infty)\), at \(x = 2\) there is a relative maximum. \(y(2)=-\frac{2^{3}}{3}+2^{2}=-\frac{8}{3}+4=\frac{4}{3}\).

Step3: Find the second - derivative \(y''\)

Differentiate \(y'=-x^{2}+2x\) using the power rule. \(y''=-2x + 2\).
Set \(y'' = 0\) for inflection points: \(-2x+2=0\), solve for \(x\): \(x = 1\).

• Test intervals for concavity:
• Choose a test point in \((-\infty,1)\), say \(x = 0\). Then \(y''=-2\times0+2=2>0\), so the function is concave up on \((-\infty,1)\).
• Choose a test point in \((1,\infty)\), say \(x = 2\). Then \(y''=-2\times2+2=-2<0\), so the function is concave down on \((1,\infty)\).

Answer:

• \(x\) - intercepts: \(x = 0\) and \(x = 3\)
• \(y\) - intercept: \(y = 0\)
• Critical points: \(x = 0\) (relative minimum) and \(x = 2\) (relative maximum)
• Increasing interval: \((0,2)\)
• Decreasing intervals: \((-\infty,0)\) and \((2,\infty)\)
• Inflection point: \(x = 1\)
• Concave up interval: \((-\infty,1)\)
• Concave down interval: \((1,\infty)\)
• Relative minimum: \((0,0)\)
• Relative maximum: \((2,\frac{4}{3})\)