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 \(y^{\prime}=-x^{2}+2x=-x(x - 2)\).

Step3: Find critical points

Set \(y^{\prime}=0\), so \(-x(x - 2)=0\). Solving for \(x\), we get \(x = 0\) and \(x = 2\).

Step4: Determine intervals of increase and decrease

Use a sign - chart for \(y^{\prime}\).

• Choose test points: For \(x\lt0\) (e.g., \(x=-1\)), \(y^{\prime}=-(-1)(-1 - 2)=-3\lt0\).
• For \(0\lt x\lt2\) (e.g., \(x = 1\)), \(y^{\prime}=-(1)(1 - 2)=1\gt0\).
• For \(x\gt2\) (e.g., \(x = 3\)), \(y^{\prime}=-(3)(3 - 2)=-3\lt0\).

The function is decreasing on \((-\infty,0)\cup(2,\infty)\) and increasing on \((0,2)\).

Step5: Find relative minima and maxima

Using the first - derivative test:

• At \(x = 0\), since the function changes from decreasing (\(x\lt0\)) to increasing (\(0\lt x\lt2\)), \(y(0)=0\) is a relative minimum.
• At \(x = 2\), since the function changes from increasing (\(0\lt x\lt2\)) to decreasing (\(x\gt2\)), \(y(2)=-\frac{8}{3}+4=\frac{4}{3}\) is a relative maximum.

Step6: Find the second - derivative

Differentiate \(y^{\prime}=-x^{2}+2x\). \(y^{\prime\prime}=-2x + 2=-2(x - 1)\).

Step7: Find inflection points

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

Step8: Analyze concavity

• For \(x\lt1\) (e.g., \(x = 0\)), \(y^{\prime\prime}=-2(0 - 1)=2\gt0\), the function is concave up on \((-\infty,1)\).
• For \(x\gt1\) (e.g., \(x = 2\)), \(y^{\prime\prime}=-2(2 - 1)=-2\lt0\), 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 \((0,0)\), relative minimum \((0,0)\), relative maximum \((2,\frac{4}{3})\), and 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 \((1,\infty)\).