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

Step2: Find the first - derivative

Differentiate \(y =-\frac{x^{3}}{3}+x^{2}\) using the power rule \(y'=-x^{2}+2x=-x(x - 2)\)

Step3: Find critical points

Set \(y'=0\). Then \(-x(x - 2)=0\), which gives \(x = 0\) and \(x = 2\)

Step4: Determine intervals of increase and decrease

• For \(x<0\), let \(x=-1\). Then \(y'=-(-1)^{2}+2(-1)=-1 - 2=-3<0\). So the function is decreasing on \((-\infty,0)\)
• For \(00\). So the function is increasing on \((0,2)\)
• For \(x>2\), let \(x = 3\). Then \(y'=-3^{2}+2\times3=-9 + 6=-3<0\). So the function is decreasing on \((2,\infty)\)

Step5: Find relative minima and maxima

• Using the first - derivative test:
• At \(x = 0\), since the function changes from decreasing \((x<0)\) to increasing \((0<x<2)\) is incorrect (it should be from decreasing \((x < 0)\) to increasing \((0<x<2)\) is wrong, actually at \(x = 0\), \(y(0)=0\) and since the function changes from decreasing \((x<0)\) to increasing \((0<x<2)\) is wrong. Wait, no: when \(x<0,y'<0\) (decreasing), when \(00\) (increasing). So \(x = 0\) is a relative minimum. \(y(0)=0\)
• At \(x = 2\), since the function changes from increasing \((0<x<2)\) to decreasing \((x>2)\), \(y(2)=-\frac{2^{3}}{3}+2^{2}=-\frac{8}{3}+4=\frac{4}{3}\). So \(x = 2\) is a relative maximum

Step6: Find the second - derivative

Differentiate \(y'=-x^{2}+2x\) using the power rule. \(y''=-2x + 2=-2(x - 1)\)

Step7: Find inflection points

Set \(y''=0\). 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: Determine concavity

• For \(x<1\), let \(x = 0\). Then \(y''=-2\times0+2=2>0\). So the function is concave up on \((-\infty,1)\)
• For \(x>1\), let \(x = 2\). Then \(y''=-2\times2+2=-2<0\). So the function is concave down on \((1,\infty)\)

Answer:

• \(y -\)intercept: \((0,0)\)
• Critical points: \(x = 0\) (relative minimum, \(y(0)=0\)) and \(x = 2\) (relative maximum, \(y(2)=\frac{4}{3}\))
• 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 information about intervals of increase/decrease and concavity to draw the curve. The function is decreasing on \((-\infty,0)\), increasing on \((0,2)\) and decreasing on \((2,\infty)\). It is concave up on \((-\infty,1)\) and concave down on \((1,\infty)\)