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 \(x = 2\).

• When \(x\lt0\), let \(x=-1\), \(y^\prime=-(-1)^{2}+2(-1)=-1 - 2=-3\lt0\), the function is decreasing.
• When \(0\lt x\lt2\), let \(x = 1\), \(y^\prime=-1^{2}+2\times1=1\gt0\), the function is increasing.
• When \(x\gt2\), let \(x = 3\), \(y^\prime=-3^{2}+2\times3=-9 + 6=-3\lt0\), the function is decreasing.

At \(x = 0\), \(y=\frac{-0^{3}}{3}+0^{2}=0\). Since the function changes from decreasing (\(x\lt0\)) to increasing (\(0\lt x\lt2\)), \(x = 0\) is not a relative extremum.
At \(x = 2\), \(y=\frac{-2^{3}}{3}+2^{2}=\frac{-8}{3}+4=\frac{4}{3}\). Since the function changes from increasing (\(0\lt x\lt2\)) to decreasing (\(x\gt2\)), \(x = 2\) is a relative maximum.

Step3: Find the second - derivative

Differentiate \(y^\prime=-x^{2}+2x\) to get \(y^{\prime\prime}=-2x + 2\). Set \(y^{\prime\prime}=0\), then \(-2x+2=0\), \(x = 1\).

• When \(x\lt1\), let \(x = 0\), \(y^{\prime\prime}=-2\times0+2=2\gt0\), the function is concave up.
• When \(x\gt1\), let \(x = 2\), \(y^{\prime\prime}=-2\times2+2=-2\lt0\), the function is concave down.

At \(x = 1\), \(y=\frac{-1^{3}}{3}+1^{2}=\frac{-1}{3}+1=\frac{2}{3}\). So the inflection point is \((1,\frac{2}{3})\).

Answer:

• \(x\) - intercepts: \((0,0)\) and \((3,0)\)
• Critical points: \(x = 0\) (not a relative extremum), \(x = 2\) (relative maximum at \((2,\frac{4}{3})\))
• Increasing interval: \((0,2)\)
• Decreasing intervals: \((-\infty,0)\) and \((2,\infty)\)
• Inflection point: \((1,\frac{2}{3})\)
• Concave up interval: \((-\infty,1)\)
• Concave down interval: \((1,\infty)\)

The graph has \(x\) - intercepts at \((0,0)\) and \((3,0)\), a relative maximum at \((2,\frac{4}{3})\), and an inflection point at \((1,\frac{2}{3})\). It is decreasing on \((-\infty,0)\) and \((2,\infty)\), increasing on \((0,2)\), concave up on \((-\infty,1)\) and concave down on \((1,\infty)\).