Difference between revisions of "009A Sample Final 3, Problem 6"
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| − | | | + | |We start by taking the derivative of <math style="vertical-align: -5px">f(x).</math> We have <math style="vertical-align: -5px">f'(x)=24x^2-4x^3.</math> |
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| − | | | + | |Now, we set <math style="vertical-align: -5px">f'(x)=0.</math> So, we have <math style="vertical-align: -6px">0=4x^2(6-x).</math> |
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| − | | | + | |Hence, we have <math style="vertical-align: 0px">x=0</math> and <math style="vertical-align: -1px">x=6.</math> |
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| − | | | + | |So, these values of <math style="vertical-align: 0px">x</math> break up the number line into 3 intervals: <math style="vertical-align: -5px">(-\infty,0),(0,6),(6,\infty).</math> |
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Revision as of 20:57, 6 March 2017
Let
(a) Over what -intervals is increasing/decreasing?
(b) Find all critical points of and test each for local maximum and local minimum.
(c) Over what -intervals is concave up/down?
(d) Sketch the shape of the graph of
| Foundations: |
|---|
| 1. is increasing when and is decreasing when |
| 2. The First Derivative Test tells us when we have a local maximum or local minimum. |
| 3. is concave up when and is concave down when |
Solution:
(a)
| Step 1: |
|---|
| We start by taking the derivative of We have |
| Now, we set So, we have |
| Hence, we have and |
| So, these values of break up the number line into 3 intervals: |
| Step 2: |
|---|
(b)
| Step 1: |
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| Step 2: |
|---|
(c)
| Step 1: |
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| Step 2: |
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| (d): |
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| Insert graph |
| Final Answer: |
|---|
| (a) |
| (b) |
| (c) |
| (d) See above |