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.
y = -\frac{x^4}{4} + x^2

Explanation:

Step1: Find the \(x -\)intercepts

Set \(y = 0\), then \(\frac{-x^{3}}{3}+x^{2}=0\). Factor out \(-\frac{x^{2}}{3}\), we get \(-\frac{x^{2}}{3}(x - 3)=0\).
So \(x = 0\) or \(x=3\).

Step2: Find the first - derivative

Differentiate \(y=\frac{-x^{3}}{3}+x^{2}\) using the power rule \(y^\prime=-x^{2}+2x\).
Set \(y^\prime = 0\), then \(-x^{2}+2x=0\). Factor out \(-x\), we have \(-x(x - 2)=0\). So \(x = 0\) or \(x = 2\).

Step3: Determine the intervals of increase and decrease

We use test points.
For the interval \((-\infty,0)\), let \(x=-1\), then \(y^\prime=-(-1)^{2}+2(-1)=-3<0\).
For the interval \((0,2)\), let \(x = 1\), then \(y^\prime=-1^{2}+2\times1 = 1>0\).
For the interval \((2,\infty)\), let \(x = 3\), then \(y^\prime=-3^{2}+2\times3=-3<0\).
So the function is decreasing on \((-\infty,0)\cup(2,\infty)\) and increasing on \((0,2)\).

Step4: Find the relative minima and maxima

Since the function changes from decreasing to increasing at \(x = 0\), \(y(0)=\frac{-0^{3}}{3}+0^{2}=0\) (a relative minimum).
Since the function changes from increasing to decreasing at \(x = 2\), \(y(2)=\frac{-2^{3}}{3}+2^{2}=\frac{-8 + 12}{3}=\frac{4}{3}\) (a relative maximum).

Step5: Find the second - derivative

Differentiate \(y^\prime=-x^{2}+2x\) to get \(y^{\prime\prime}=-2x + 2\).
Set \(y^{\prime\prime}=0\), then \(-2x+2 = 0\), so \(x = 1\).
When \(x = 1\), \(y(1)=\frac{-1^{3}}{3}+1^{2}=\frac{2}{3}\). So the inflection point is \((1,\frac{2}{3})\).

Step6: Determine the concavity

For the interval \((-\infty,1)\), let \(x = 0\), then \(y^{\prime\prime}=-2\times0+2=2>0\) (concave up).
For the interval \((1,\infty)\), let \(x = 2\), then \(y^{\prime\prime}=-2\times2+2=-2<0\) (concave down).

Answer:

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