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 critical points

Set \(y^\prime = 0\). So \(-x(x - 2)=0\). Solving \( - x(x - 2)=0\) gives \(x = 0\) and \(x = 2\).

Step4: Determine intervals of increase and decrease

• For the interval \((-\infty,0)\), let \(x=-1\). Then \(y^\prime=-(-1)(-1 - 2)=-3<0\). So the function is decreasing on \((-\infty,0)\).
• For the interval \((0,2)\), let \(x = 1\). Then \(y^\prime=-1(1 - 2)=1>0\). 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 \((2,\infty)\).

Step5: Find 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 there is a relative minimum at \((0,0)\).
• 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 there is a relative maximum at \((2,\frac{4}{3})\).

Step6: Find the second - derivative

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

Step7: Find inflection points

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: Analyze concavity

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

Answer:

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

To 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 and concavity to draw the curve. The function is decreasing from \((-\infty,0)\), increasing from \((0,2)\), and decreasing from \((2,\infty)\). It is concave up from \((-\infty,1)\) and concave down from \((1,\infty)\).