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, 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 \(y=0\).

Step2: Find critical points

Differentiate \(y\) using the power rule. \(y^{\prime}=-x^{2}+2x\). Set \(y^{\prime}=0\), so \(-x^{2}+2x = 0\). Factor out \(-x\): \(-x(x - 2)=0\). The critical points are \(x = 0\) and \(x = 2\).

Step3: Determine intervals of increase and decrease

Test intervals using the first - derivative.

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

Step4: Find relative minima and maxima

Use 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 \(x = 0\), \(y^{\prime}\) changes from negative (left of \(x = 0\)) to positive (right of \(x = 0\)) is wrong. Wait, no: when \(x<0,y^{\prime}<0\); when \(00\). So \(x = 0\) is a relative minimum. \(y(0)=0\).
• At \(x = 2\), since the function changes from increasing \((0<x<2)\) to decreasing \((x>2)\), \(x = 2\) is a relative maximum. \(y(2)=-\frac{8}{3}+4=\frac{-8 + 12}{3}=\frac{4}{3}\).

Step5: Find inflection points

Differentiate \(y^{\prime}=-x^{2}+2x\) to get \(y^{\prime\prime}=-2x + 2\). Set \(y^{\prime\prime}=0\), then \(-2x+2 = 0\), \(x = 1\). When \(x = 1\), \(y=-\frac{1}{3}+1=\frac{2}{3}\).

Step6: Determine concavity

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

Answer:

• \(y-\)intercept: \((0,0)\)
• Critical points: \(x = 0\) (relative minimum, \(y = 0\)) and \(x = 2\) (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 \((2,\infty)\). It is concave up on \((-\infty,1)\) and concave down on \((1,\infty)\).