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

Step2: Find critical points

First, find the first derivative \(y^{\prime}\). Using the power rule \(y^{\prime}=-x^{2}+2x\).
Set \(y^{\prime}=0\), so \(-x^{2}+2x = 0\). Factor out \(-x\): \(-x(x - 2)=0\).
Solving for \(x\), we get \(x = 0\) and \(x = 2\).

Step3: Determine intervals of increase and decrease

We use test intervals. The critical points divide the real - line into intervals \((-\infty,0)\), \((0,2)\) and \((2,\infty)\).
For the interval \((-\infty,0)\), let \(x=-1\). Then \(y^{\prime}=-(-1)^{2}+2(-1)=-1 - 2=-3<0\). So the function is decreasing on \((-\infty,0)\).
For the interval \((0,2)\), let \(x = 1\). Then \(y^{\prime}=-1^{2}+2\times1=1>0\). So the function is increasing on \((0,2)\).
For the interval \((2,\infty)\), let \(x = 3\). Then \(y^{\prime}=-3^{2}+2\times3=-9 + 6=-3<0\). 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 \(x = 2\), \(y(2)=-\frac{2^{3}}{3}+2^{2}=-\frac{8}{3}+4=\frac{4}{3}\) is a relative maximum.

Step4: Find inflection points

Find the second derivative \(y^{\prime\prime}\). Differentiating \(y^{\prime}=-x^{2}+2x\), we get \(y^{\prime\prime}=-2x + 2\).
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 \((1,\frac{2}{3})\).
For \(x<1\) (e.g., \(x = 0\)), \(y^{\prime\prime}=-2\times0+2=2>0\), the function is concave up on \((-\infty,1)\).
For \(x>1\) (e.g., \(x = 2\)), \(y^{\prime\prime}=-2\times2+2=-2<0\), the function is concave down on \((1,\infty)\).

Answer:

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