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Math 112 Solutions for 3.3, p. 248

1.
(b) global minimum at . (f) global maximum at . (j) global maximum at . (n) global minimum at the origin.
2.
(b) global maximum at , global minimum at , and local minimum at .

(c) global maximum of at , global minimum of at .

(f) local maximum at , local minimum at .

(j) global minimum at .

(n) no extrema

3.
The function Is defined on but has no maximum there.

4.
changes sign at , so changes sign at and hence has an extremum.

5.
Theorem. If and , then has an inflection point at .

Proof. If , then is a critical point of at which the derivative is zero, so we may apply the second derivative test to . If , then has an extremum at , so that the derivative of changes at and hence the concavity changes at . Thus is an inflection point of .

6.
Since is continuous on a closed interval, it has a global maximum there, which is also a local maximum, so occurs at .

7.
Suppose is defined in and is the only critical point of , and has a local maximum at (the case of minimum is similar). Since is continuous and is the only critical point, is differentiable in except possibly at . If were to have a value greater than somewhere, then it must have the same value at some point, and then Rolle's Theorem would give us another critical point, which we can't have. Thus the value at is a global maximum.

8.
(a) P2 should maximize with respect to , getting .

(b) P1 should replace in with the value that P2 is going to use, and minimize the resulting , getting .

(c) P1 should minimize with respect to , getting .

(d) P2 should replace in with the value that P1 is going to use and maximize the resulting , getting .

(e) Each player has to go first" so P1 chooses and P2 chooses . (The resulting score is 0, so the game is uninteresting, even though it is fair.)

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Jason Grout 2003-02-25