© 1997 by Karl Hahn
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This problem is merely a test of your ability to deal with endpoints.
Here is a graph of
f(x) = x2 - 4x + 4, and its derivative,
f'(x) = 2x - 4.
The left-hand endpoint is at x = 2.
The right-hand endpoint is at x = 6.
We can see that f'(x) has no discontinuities, so there
ought not be any cusps in this function.
If we make a table of the values of f(x) and f'(x) at the endpoints, we get:
f(2) = 0 f'(2) = 0
f(6) = 16 f'(6) = 8
Here again is the decision tree for deciding whether endpoints are
maximums or minimums:
Decision Tree for Endpoints
left end right end
| |
| find f'(endpoint) |
/|\ /|\
/ | \ / | \
/ | \ / | \
positive zero negative positive zero negative
it's a min | it's a max it's a max | it's a min
| |
| if f'(endpoint) = 0, |
| take higher derivatives |
| at the endpoint until |
| you get one that's not |
| equal to zero |
/ \ / \
/ \ / \
/ \ / \
positive negative positive negative
it's a min it's a max / \ / \
/ \ / \
/ \ / \
derivative number even or odd? even odd even odd
it's a min it's a max it's a min
Note that our left-hand endpoint yields zero for its derivative. According to the decision tree, we take the next higher derivative until we get a nonzero value. In this case, f"(x) = 2, which is nonzero everywhere. That means that the second derivative at the left-hand endpoint is positive. The decision tree tells us that this endpoint, (2, 0), is a minimum.
The right-hand endpoint is no problem. We see that f'(6) = 8, which is positive. The decision tree tells us that this endpoint, (6, 8), a maximum.
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