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 where it is 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

Use the power rule \((x^{n})^\prime=nx^{n - 1}\). If \(y=-\frac{x^{3}}{3}+x^{2}\), then \(y^\prime=-x^{2}+2x=-x(x - 2)\).

Step3: Find the critical points

Set \(y^\prime = 0\). So \(-x(x - 2)=0\), which gives \(x = 0\) or \(x = 2\).

Step4: Determine the intervals of increase and decrease

Use a sign - chart for \(y^\prime=-x(x - 2)\).

• Choose a test point in the interval \((-\infty,0)\), say \(x=-1\). Then \(y^\prime=-(-1)(-1 - 2)=-3<0\). So \(y\) is decreasing on \((-\infty,0)\).
• Choose a test point in the interval \((0,2)\), say \(x = 1\). Then \(y^\prime=-1(1 - 2)=1>0\). So \(y\) is increasing on \((0,2)\).
• Choose a test point in the interval \((2,\infty)\), say \(x = 3\). Then \(y^\prime=-3(3 - 2)=-3<0\). So \(y\) is decreasing on \((2,\infty)\).

Step5: Find the relative minima and maxima

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

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

Step8: Determine the concavity

• Choose a test point in the interval \((-\infty,1)\), say \(x = 0\). Then \(y^{\prime\prime}=-2(0 - 1)=2>0\). So \(y\) is concave up on \((-\infty,1)\).
• Choose a test point in the interval \((1,\infty)\), say \(x = 2\). Then \(y^{\prime\prime}=-2(2 - 1)=-2<0\). So \(y\) 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 intervals of increase/decrease and concavity to draw the curve. For example, the function is decreasing on \((-\infty,0)\), increasing on \((0,2)\), and decreasing on \((2,\infty)\); concave up on \((-\infty,1)\) and concave down on \((1,\infty)\).