QUESTION IMAGE
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 )
Step1: Find \(x\) - intercept
Set \(y = 0\), then \(\frac{-x^{3}}{3}+x^{2}=0\). Factor out \(-\frac{x^{2}}{3}\), we get \(-\frac{x^{2}}{3}(x - 3)=0\). So \(x = 0\) or \(x=3\).
Step2: Find the first - derivative
\(y'=-x^{2}+2x=-x(x - 2)\)
Set \(y'=0\), then \(-x(x - 2)=0\). So \(x = 0\) or \(x = 2\)
Step3: Determine intervals of increase and decrease
- For \(y'=-x(x - 2)\), consider the intervals:
- When \(x<0\), let \(x=-1\), then \(y'=-(-1)(-1 - 2)=-3<0\). So the function is decreasing on \((-\infty,0)\)
- When \(0
- When \(x>2\), let \(x = 3\), then \(y'=-3(3 - 2)=-3<0\). So the function is decreasing on \((2,\infty)\)
Step4: Find relative extrema
- By the first - derivative test:
- At \(x = 0\): Since the function changes from decreasing \((x<0)\) to increasing \((0<x<2)\) is incorrect (it changes from decreasing \((x<0)\) to increasing \((0<x<2)\) at \(x = 0\), but \(y(0)=0\).
- At \(x = 2\): Since the function changes from increasing \((0<x<2)\) to decreasing \((x>2)\). \(y(2)=\frac{-8}{3}+4=\frac{4}{3}\). So the relative maximum is at the point \((2,\frac{4}{3})\)
Step5: Find the second - derivative
\(y''=-2x + 2=-2(x - 1)\)
Set \(y''=0\), then \(x = 1\)
Step6: Determine concavity and inflection point
- When \(x<1\), let \(x=0\), then \(y''=-2(0 - 1)=2>0\). So the function is concave up on \((-\infty,1)\)
- When \(x>1\), let \(x = 2\), then \(y''=-2(2 - 1)=-2<0\). So the function is concave down on \((1,\infty)\)
- At \(x = 1\), \(y(1)=\frac{-1}{3}+1=\frac{2}{3}\). So the inflection point is \((1,\frac{2}{3})\)
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- \(x\) - intercepts: \(x = 0\) and \(x = 3\)
- Critical points: \(x = 0\) and \(x = 2\)
- Intervals of increase: \((0,2)\)
- Intervals of decrease: \((-\infty,0)\cup(2,\infty)\)
- Relative maximum: \((2,\frac{4}{3})\)
- Inflection point: \((1,\frac{2}{3})\)
- Intervals of concave up: \((-\infty,1)\)
- Intervals of concave down: \((1,\infty)\)