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 where it is 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 \(y^{\prime}=-\frac{3x^{2}}{3}+2x=-x^{2}+2x=-x(x - 2)\)

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

• For the interval \((-\infty,0)\), let \(x=-1\). Then \(y^{\prime}=-(-1)(-1 - 2)=-3<0\), so the function is decreasing on \((-\infty,0)\)
• For the interval \((0,2)\), let \(x = 1\). Then \(y^{\prime}=-1(1 - 2)=1>0\), so the function is increasing on \((0,2)\)
• For the interval \((2,\infty)\), let \(x = 3\). Then \(y^{\prime}=-3(3 - 2)=-3<0\), so the function is decreasing on \((2,\infty)\)

Step5: Find relative minima and maxima

• Since the function changes from decreasing \((-\infty,0)\) to increasing \((0,2)\) at \(x = 0\), \(y(0)=0\) is a relative minimum.
• Since the function changes from increasing \((0,2)\) to decreasing \((2,\infty)\) at \(x = 2\), \(y(2)=-\frac{8}{3}+4=\frac{4}{3}\) 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: Determine concavity

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

Answer:

• \(y -\)intercept: \((0,0)\)
• Critical points: \(x = 0\) and \(x = 2\)
• Increasing on \((0,2)\)
• Decreasing on \((-\infty,0)\cup(2,\infty)\)
• Relative minimum at \((0,0)\)
• Relative maximum at \((2,\frac{4}{3})\)
• Inflection point at \((1,\frac{2}{3})\)
• Concave up on \((-\infty,1)\)
• Concave down on \((1,\infty)\)