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 \(y = 0\), so \(\frac{-x^{3}}{3}+x^{2}=0\). Factor out \(-\frac{x^{2}}{3}\), we get \(-\frac{x^{2}}{3}(x - 3)=0\).
Solving \(-\frac{x^{2}}{3}(x - 3)=0\) gives \(x = 0\) 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 \(-x(x - 2)=0\).
Solving \(-x(x - 2)=0\) gives \(x = 0\) or \(x = 2\).

• For \(x<0\), let \(x=-1\), then \(y^\prime=-(-1)^{2}+2(-1)=-1 - 2=-3<0\).
• For \(00\).
• For \(x>2\), let \(x = 3\), then \(y^\prime=-3^{2}+2\times3=-9 + 6=-3<0\).

The function is decreasing on \((-\infty,0)\cup(2,\infty)\) and increasing on \((0,2)\).
At \(x = 0\), \(y=\frac{-0^{3}}{3}+0^{2}=0\). At \(x = 2\), \(y=\frac{-2^{3}}{3}+2^{2}=\frac{-8 + 12}{3}=\frac{4}{3}\). So the relative minimum is at \((0,0)\) and the relative maximum is at \((2,\frac{4}{3})\).

Step3: Find the second - derivative

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

• For \(x<1\), let \(x = 0\), then \(y^{\prime\prime}=-2\times0+2=2>0\).
• For \(x>1\), let \(x = 2\), then \(y^{\prime\prime}=-2\times2+2=-2<0\).

The inflection point is at \(x = 1\). When \(x = 1\), \(y=\frac{-1^{3}}{3}+1^{2}=\frac{-1 + 3}{3}=\frac{2}{3}\). The function is concave up on \((-\infty,1)\) and concave down on \((1,\infty)\).

Answer:

• \(x -\)intercepts: \((0,0)\) and \((3,0)\)
• Critical points: \((0,0)\) (relative minimum) and \((2,\frac{4}{3})\) (relative maximum)
• Function is decreasing on \((-\infty,0)\cup(2,\infty)\) and increasing on \((0,2)\)
• Inflection point: \((1,\frac{2}{3})\)
• Function is concave up on \((-\infty,1)\) and concave down on \((1,\infty)\)