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^3

Explanation:

Step1: Find \(y -\)intercept

Set \(x = 0\) in \(y=-\frac{x^{3}}{3}+x^{2}\). Then \(y=0\). So \(y -\)intercept is \((0,0)\).

Step2: Find critical points

Differentiate \(y\) with respect to \(x\). \(y^{\prime}=-x^{2}+2x=-x(x - 2)\).
Set \(y^{\prime}=0\), then \(-x(x - 2)=0\). Solving gives \(x = 0\) and \(x = 2\).

Step3: Determine intervals of increase and decrease

Use test - points.
For the interval \((-\infty,0)\), let \(x=-1\). Then \(y^{\prime}=-(-1)^{2}+2(-1)=-3<0\), so \(y\) is decreasing on \((-\infty,0)\).
For the interval \((0,2)\), let \(x = 1\). Then \(y^{\prime}=-1^{2}+2\times1 = 1>0\), so \(y\) is increasing on \((0,2)\).
For the interval \((2,\infty)\), let \(x = 3\). Then \(y^{\prime}=-3^{2}+2\times3=-3<0\), so \(y\) is decreasing on \((2,\infty)\).
Since \(y\) changes from decreasing to increasing at \(x = 0\), \(y(0)=0\) is a relative minimum.
Since \(y\) changes from increasing to decreasing at \(x = 2\), \(y(2)=-\frac{8}{3}+4=\frac{4}{3}\) is a relative maximum.

Step4: Find inflection points

Differentiate \(y^{\prime}\) to get \(y^{\prime\prime}=-2x + 2\).
Set \(y^{\prime\prime}=0\), then \(-2x + 2=0\), \(x = 1\).
When \(x = 1\), \(y=-\frac{1}{3}+1=\frac{2}{3}\). So the inflection point is \((1,\frac{2}{3})\).
For \(x<1\), let \(x = 0\), \(y^{\prime\prime}=2>0\), the function is concave up on \((-\infty,1)\).
For \(x>1\), let \(x = 2\), \(y^{\prime\prime}=-2<0\), 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 on \((0,2)\), decreasing on \((-\infty,0)\cup(2,\infty)\)
• Inflection point: \((1,\frac{2}{3})\)
• Concave up on \((-\infty,1)\), concave down on \((1,\infty)\)