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 \(y^{\prime}=x(-x + 2)\).
Set \(y^{\prime}=0\), then \(x(-x + 2)=0\). Solving \(x(-x + 2)=0\) gives \(x = 0\) or \(x = 2\).

Step3: Determine intervals of increase and decrease

Use a test - point method.

• For the interval \((-\infty,0)\), let \(x=-1\). Then \(y^{\prime}=(-1)(-(-1)+2)=(-1)(-1 + 2)=-1<0\). So the function is decreasing on \((-\infty,0)\).
• For the interval \((0,2)\), let \(x = 1\). Then \(y^{\prime}=(1)(-1 + 2)=1>0\). So the function is increasing on \((0,2)\).
• For the interval \((2,\infty)\), let \(x = 3\). Then \(y^{\prime}=(3)(-3 + 2)=-3<0\). 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 \(x = 2\), \(y(2)=-\frac{8}{3}+4=\frac{4}{3}\) is a relative maximum.

Step4: Find the second - derivative

Differentiate \(y^{\prime}=-x^{2}+2x\) using the power rule. \(y^{\prime\prime}=-2x + 2\).
Set \(y^{\prime\prime}=0\), then \(-2x + 2=0\), which gives \(x = 1\).
When \(x = 1\), \(y=-\frac{1}{3}+1=\frac{2}{3}\). So the inflection point is \((1,\frac{2}{3})\).

• For \(x<1\) (e.g., \(x = 0\)), \(y^{\prime\prime}=2>0\), the function is concave up on \((-\infty,1)\).
• For \(x>1\) (e.g., \(x = 2\)), \(y^{\prime\prime}=-2<0\), the function is concave down on \((1,\infty)\).

Step5: Sketch the graph

• Plot the \(y -\)intercept \((0,0)\), relative minimum \((0,0)\), relative maximum \((2,\frac{4}{3})\), and inflection point \((1,\frac{2}{3})\).
• Use the intervals of increase/decrease (decreasing on \((-\infty,0)\), increasing on \((0,2)\), decreasing on \((2,\infty)\)) and concavity (concave up on \((-\infty,1)\), concave down on \((1,\infty)\)) to sketch the curve.

Answer:

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