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, 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
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 \(y'=-x^{2}+2x\).
Set \(y'=0\): \(-x^{2}+2x = 0\), factor out \(-x\) gives \(-x(x - 2)=0\). So \(x = 0\) or \(x = 2\).
Step3: Determine intervals of increase and decrease
Choose test points:
For \(x=-1\), \(y'=-(-1)^{2}+2(-1)=-1 - 2=-3<0\).
For \(x = 1\), \(y'=-1^{2}+2\times1=1>0\).
For \(x = 3\), \(y'=-3^{2}+2\times3=-9 + 6=-3<0\).
The function is increasing on the interval \((0,2)\) (since \(y'>0\) when \(0
Step4: Find relative extrema
Using the first - derivative test:
When \(x = 0\), since the function changes from decreasing (\(x<0\)) to increasing (\(0
Step5: Find the second - derivative
Differentiate \(y'=-x^{2}+2x\) to get \(y''=-2x + 2\).
Set \(y''=0\): \(-2x+2 = 0\), then \(x = 1\).
Step6: Determine concavity
For \(x<1\), let \(x = 0\), \(y''=-2\times0 + 2=2>0\), the function is concave up on \((-\infty,1)\).
For \(x>1\), let \(x = 2\), \(y''=-2\times2+2=-2<0\), the function is concave down on \((1,\infty)\).
The inflection point is at \(x = 1\), \(y(1)=-\frac{1^{3}}{3}+1^{2}=\frac{2}{3}\), so the inflection point is \((1,\frac{2}{3})\).
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- \(y -\)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: \((1,\infty)\)