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 \(y -\)intercept

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

Step2: Find critical points

First, find the first - derivative \(y^{\prime}=-x^{2}+2x\). Set \(y^{\prime}=0\), then \(-x^{2}+2x = 0\), \(x(-x + 2)=0\). So \(x = 0\) or \(x = 2\).

Step3: Determine intervals of increase and decrease

Use the test - point method.
For the interval \((-\infty,0)\), let \(x=-1\), then \(y^{\prime}=-(-1)^{2}+2(-1)=-3<0\). 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 \(y^{\prime}=-3^{2}+2\times3=-3<0\). So the function is decreasing on \((2,\infty)\).
Since the function changes from decreasing to increasing at \(x = 0\), \(y(0)=0\) is a relative minimum. Since the function changes from increasing to decreasing at \(x = 2\), \(y(2)=-\frac{8}{3}+4=\frac{4}{3}\) is a relative maximum.

Step4: Find inflection points

Find the second - derivative \(y^{\prime\prime}=-2x + 2\). Set \(y^{\prime\prime}=0\), then \(-2x+2 = 0\), \(x = 1\). When \(x = 1\), \(y=-\frac{1}{3}+1=\frac{2}{3}\). So the inflection point is \((1,\frac{2}{3})\).
For the interval \((-\infty,1)\), let \(x = 0\), \(y^{\prime\prime}=2>0\), the function is concave up on \((-\infty,1)\).
For the interval \((1,\infty)\), let \(x = 2\), \(y^{\prime\prime}=-2<0\), the function is concave down on \((1,\infty)\).

Sketch:

• 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 (decreasing on \((-\infty,0)\), increasing on \((0,2)\), decreasing on \((2,\infty)\)) and concavity (concave up on \((-\infty,1)\), concave down on \((1,\infty)\)) to sketch the curve.

Answer:

• \(y -\)intercept: \((0,0)\)
• Critical points: \(x = 0\) (relative minimum) and \(x = 2\) (relative maximum)
• 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: \((1,\infty)\)