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 \((x^{n})^\prime=nx^{n - 1}\).
\(y^\prime=-x^{2}+2x=-x(x - 2)\)
Step3: Find the critical points
Set \(y^\prime=0\), so \(-x(x - 2)=0\).
Solving \(-x(x - 2)=0\) gives \(x = 0\) and \(x = 2\)
Step4: Determine intervals of increase and decrease
Use a sign chart for \(y^\prime=-x(x - 2)\).
- Choose test points: For \(x=-1\), \(y^\prime=-(-1)(-1 - 2)=-3<0\)
- For \(x = 1\), \(y^\prime=-1(1 - 2)=1>0\)
- For \(x = 3\), \(y^\prime=-3(3 - 2)=-3<0\)
The function is increasing on the interval \((0,2)\) (since \(y^\prime>0\) when \(0
Step5: Find the second - derivative
Differentiate \(y^\prime=-x^{2}+2x\) using the power rule. \(y^{\prime\prime}=-2x + 2=-2(x - 1)\)
Step6: Find the inflection point
Set \(y^{\prime\prime}=0\), then \(-2(x - 1)=0\), which gives \(x = 1\)
When \(x = 1\), \(y=-\frac{1^{3}}{3}+1^{2}=\frac{2}{3}\). So the inflection point is \((1,\frac{2}{3})\)
Step7: Find relative minima and maxima
Use the first - derivative test.
- At \(x = 0\): Since the function changes from decreasing (\(x<0,y^\prime<0\)) to increasing (\(0
0\)), \(y(0)=0\) is a relative minimum. - At \(x = 2\): Since the function changes from increasing (\(0
0\)) to decreasing (\(x>2,y^\prime<0\)), \(y(2)=-\frac{2^{3}}{3}+2^{2}=-\frac{8}{3}+4=\frac{4}{3}\) is a relative maximum
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- \(y -\)intercept: \((0,0)\)
- Critical points: \(x = 0\) and \(x = 2\)
- Increasing interval: \((0,2)\)
- Decreasing intervals: \((-\infty,0)\) and \((2,\infty)\)
- Inflection point: \((1,\frac{2}{3})\)
- Relative minimum: \((0,0)\)
- Relative maximum: \((2,\frac{4}{3})\)
To sketch the graph:
- Plot the \(y -\)intercept \((0,0)\), inflection point \((1,\frac{2}{3})\), relative minimum \((0,0)\) and relative maximum \((2,\frac{4}{3})\)
- Use the intervals of increase and decrease to shape the curve. The function is decreasing on \((-\infty,0)\), increasing on \((0,2)\) and decreasing on \((2,\infty)\)