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 \(y -\)intercept

Set \(x = 0\) in \(y=-\frac{x^{3}}{3}+x^{2}\). Then \(y=-\frac{0^{3}}{3}+0^{2}=0\). So the \(y -\)intercept is \((0,0)\).

Step2: Find the first - derivative

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 \(-x\) gives \(-x(x - 2)=0\). So \(x = 0\) and \(x = 2\) are critical points.

Step3: Determine intervals of increase and decrease

Use a test - point method.
For the interval \((-\infty,0)\), let \(x=-1\). Then \(y'=-(-1)^{2}+2(-1)=-1 - 2=-3<0\), so the function is decreasing on \((-\infty,0)\).
For the interval \((0,2)\), let \(x = 1\). Then \(y'=-1^{2}+2\times1=1>0\), so the function is increasing on \((0,2)\).
For the interval \((2,\infty)\), let \(x = 3\). Then \(y'=-3^{2}+2\times3=-9 + 6=-3<0\), so the function is decreasing on \((2,\infty)\).
Since the function changes from decreasing to increasing at \(x = 0\), \(y(0)=0\) is a relative minimum. Since the function changes from increasing to decreasing at \(x = 2\), \(y(2)=-\frac{2^{3}}{3}+2^{2}=-\frac{8}{3}+4=\frac{4}{3}\) is a relative maximum.

Step4: Find the second - derivative

Differentiate \(y'=-x^{2}+2x\) to get \(y''=-2x + 2\).
Set \(y'' = 0\): \(-2x+2=0\), then \(x = 1\). When \(x = 1\), \(y=-\frac{1^{3}}{3}+1^{2}=\frac{2}{3}\). So the inflection point is \((1,\frac{2}{3})\).
For \(x<1\) (e.g., \(x = 0\)), \(y''=-2\times0 + 2=2>0\), the function is concave up on \((-\infty,1)\).
For \(x>1\) (e.g., \(x = 2\)), \(y''=-2\times2+2=-2<0\), the function is concave down on \((1,\infty)\).

Answer:

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