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 first - derivative

The function is \(y =-\frac{x^{3}}{3}+x^{2}\).
Using the power rule \((x^{n})^\prime=nx^{n - 1}\), the first - derivative \(y^\prime=-x^{2}+2x=-x(x - 2)\).

Step2: Find critical points

Set \(y^\prime = 0\), then \(-x(x - 2)=0\).
Solving the equation \(-x(x - 2)=0\) gives \(x = 0\) and \(x = 2\).

Step3: Determine intervals of increase and decrease

We consider the intervals \((-\infty,0)\), \((0,2)\) and \((2,\infty)\).

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

Step4: Find relative minima and maxima

Using the first - derivative test:

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

Step5: Find the second - derivative

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

Step6: Find inflection points

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

Step7: Analyze concavity

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

Step8: Find \(y\) - intercept

Set \(x = 0\) in \(y =-\frac{x^{3}}{3}+x^{2}\), we get \(y = 0\). So the \(y\) - intercept is \((0,0)\).

Step9: Sketch the graph

Plot the \(y\) - intercept \((0,0)\), critical points \((0,0)\) and \((2,\frac{4}{3})\), inflection point \((1,\frac{2}{3})\). Use the information about intervals of increase/decrease and concavity to sketch the curve.

Answer:

• \(y\) - intercept: \((0,0)\)
• Critical points: \(x = 0\) (relative minimum at \((0,0)\)) and \(x = 2\) (relative maximum at \((2,\frac{4}{3})\))
• Intervals of increase: \((0,2)\)
• Intervals of decrease: \((-\infty,0)\cup(2,\infty)\)
• Inflection point: \((1,\frac{2}{3})\)
• Interval of concave up: \((-\infty,1)\)
• Interval of concave down: \((1,\infty)\)[LLM SSE On Failure]