Difference between revisions of "1950 AHSME Problems/Problem 41"
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==Solution 2 == | ==Solution 2 == | ||
− | Using calculus, the derivative of the function is <math>2ax + b</math>, which has a critical point at <math>\frac{-b}{2a}</math>. The second derivative of the function is <math>2a</math>; since <math>a > 0</math>, this critical point is a minimum. As in Solution 1, plug this value into the function to obtain <math>\boxed{\textbf{(D)}\ \dfrac{4ac-b^2}{4a}}</math>. | + | Using calculus, we find that the derivative of the function is <math>2ax + b</math>, which has a critical point at <math>\frac{-b}{2a}</math>. The second derivative of the function is <math>2a</math>; since <math>a > 0</math>, this critical point is a minimum. As in Solution 1, plug this value into the function to obtain <math>\boxed{\textbf{(D)}\ \dfrac{4ac-b^2}{4a}}</math>. |
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+ | ~ [https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi] | ||
==See Also== | ==See Also== |
Latest revision as of 13:43, 4 April 2024
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
.
~ cxsmi
See Also
1950 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 40 |
Followed by Problem 42 | |
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All AHSME Problems and Solutions |
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