1950 AHSME Problems/Problem 41
Contents
Problem
The least value of the function with
is:
Solution
The vertex of a parabola is at for
. Because
, the vertex is a minimum. Therefore
.
Solution 2
Using calculus, we find that the derivative of the function is , which has a critical point at
. The second derivative of the function is
; since
, this critical point is a minimum. As in Solution 1, plug this value into the function to obtain
.
See Also
1950 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 40 |
Followed by Problem 42 | |
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