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 \(y=0\). So the \(y -\)intercept is \((0,0)\).

Step2: Find the first - derivative

Differentiate \(y =-\frac{x^{3}}{3}+x^{2}\) using the power rule \(y^\prime=\frac{d}{dx}(-\frac{1}{3}x^{3}+x^{2})\).
\(y^\prime=-x^{2}+2x=-x(x - 2)\)

Step3: Find critical points

Set \(y^\prime = 0\). Then \(-x(x - 2)=0\). 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 the interval \((-\infty,0)\), let \(x=-1\). Then \(y^\prime=-(-1)(-1 - 2)=-3<0\).
• For the interval \((0,2)\), let \(x = 1\). Then \(y^\prime=-1(1 - 2)=1>0\).
• For the interval \((2,\infty)\), let \(x = 3\). Then \(y^\prime=-3(3 - 2)=-3<0\).

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

Step5: Find relative minima and maxima

Use the first - derivative test.

• At \(x = 0\): Since the function changes from decreasing \((x<0)\) to increasing \((x>0)\), \(f(0)=0\) is a relative minimum.
• At \(x = 2\): Since the function changes from increasing \((x<2)\) to decreasing \((x>2)\), \(f(2)=-\frac{8}{3}+4=\frac{4}{3}\) 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 \(x<1\) (e.g., \(x = 0\)), \(y^{\prime\prime}=2>0\), the function is concave up on \((-\infty,1)\).
• For \(x>1\) (e.g., \(x = 2\)), \(y^{\prime\prime}=-2<0\), 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, \(y=\frac{4}{3}\))
• Increasing interval: \((0,2)\)
• Decreasing intervals: \((-\infty,0)\) and \((2,\infty)\)
• Inflection point: \((1,\frac{2}{3})\)
• Concave up: \((-\infty,1)\)
• Concave down: \((1,\infty)\)