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 \((x^{n})^\prime=nx^{n - 1}\).
\(y^\prime=-x^{2}+2x=-x(x - 2)\)

Step3: Find the critical points

Set \(y^\prime=0\), then \(-x(x - 2)=0\). Solving for \(x\), we get \(x = 0\) and \(x = 2\).

Step4: Determine the intervals of increase and decrease

• For the interval \((-\infty,0)\), let \(x=-1\). Then \(y^\prime=-(-1)^{2}+2(-1)=-1 - 2=-3<0\). So the function is decreasing on \((-\infty,0)\).
• For the interval \((0,2)\), let \(x = 1\). Then \(y^\prime=-1^{2}+2\times1=1>0\). So the function is increasing on \((0,2)\).
• For the interval \((2,\infty)\), let \(x = 3\). Then \(y^\prime=-3^{2}+2\times3=-9 + 6=-3<0\). So the function is decreasing on \((2,\infty)\).

Step5: Find the relative minima and maxima

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

Step6: Find the second - derivative

Differentiate \(y^\prime=-x^{2}+2x\). Using the power rule, \(y^{\prime\prime}=-2x + 2=-2(x - 1)\)

Step7: Find the inflection points

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

Step8: Determine the concavity

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

Answer:

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

To sketch the graph:

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