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 \(y=0\). 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 \(y^\prime = 0\), so \(-x(x - 2)=0\).
Solving \(-x(x - 2)=0\) gives \(x = 0\) and \(x = 2\)

Step4: Determine the intervals of increase and decrease

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

• Choose a test point in \((-\infty,0)\), say \(x=-1\). Then \(y^\prime=-(-1)(-1 - 2)=-3<0\), so the function is decreasing on \((-\infty,0)\)
• Choose a test point in \((0,2)\), say \(x = 1\). Then \(y^\prime=-1(1 - 2)=1>0\), so the function is increasing on \((0,2)\)
• Choose a test point in \((2,\infty)\), say \(x = 3\). Then \(y^\prime=-3(3 - 2)=-3<0\), 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 \(x = 0\), \(y=-\frac{0^{3}}{3}+0^{2}=0\), so \((0,0)\) is a relative minimum.
Since the function changes from increasing \((0,2)\) to decreasing \((2,\infty)\), at \(x = 2\), \(y=-\frac{2^{3}}{3}+2^{2}=-\frac{8}{3}+4=\frac{4}{3}\), so \((2,\frac{4}{3})\) 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 \(y^{\prime\prime}=0\), so \(-2(x - 1)=0\), 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 intervals of concavity

• For \(y^{\prime\prime}=-2(x - 1)\), if \(x<1\), say \(x = 0\), \(y^{\prime\prime}=-2(0 - 1)=2>0\), so the function is concave up on \((-\infty,1)\)
• If \(x>1\), say \(x = 2\), \(y^{\prime\prime}=-2(2 - 1)=-2<0\), so the function is concave down on \((1,\infty)\)

Answer:

• \(y -\)intercept: \((0,0)\)
• Critical points: \(x = 0\) and \(x = 2\)
• Intervals of increase: \((0,2)\)
• Intervals of decrease: \((-\infty,0)\cup(2,\infty)\)
• Relative minimum: \((0,0)\)
• Relative maximum: \((2,\frac{4}{3})\)
• 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. The function is a cubic function opening downwards (because the leading coefficient of \(y =-\frac{x^{3}}{3}+x^{2}\) is negative for the \(x^{3}\) term).