# Difference between revisions of "1965 AHSME Problems/Problem 34"

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\textbf{(E) }\ \frac{34}{5} </math> | \textbf{(E) }\ \frac{34}{5} </math> | ||

− | ==Solution== | + | ==Solution 1== |

To begin, lets denote the equation, <math>\frac {4x^2 + 8x + 13}{6(1 + x)}</math> as <math>f(x)</math>. Let's notice that: | To begin, lets denote the equation, <math>\frac {4x^2 + 8x + 13}{6(1 + x)}</math> as <math>f(x)</math>. Let's notice that: | ||

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<math>(\text{With equality when } \frac{2(x+1)}{3} = \frac{3}{2(x+1)}, \text{ or } x=\frac{1}{2})</math> | <math>(\text{With equality when } \frac{2(x+1)}{3} = \frac{3}{2(x+1)}, \text{ or } x=\frac{1}{2})</math> | ||

+ | |||

+ | ==Solution 2 (Calculus)== | ||

+ | Take the derivative of f(x) and f'(x) using the quotient rule. | ||

+ | <cmath>\begin{align*} | ||

+ | f(x) & = \frac {4x^2 + 8x + 13}{6(1 + x)}\\\\ | ||

+ | f'(x) & = \frac{(4x^2 + 8x + 13)'(1 + x) - (4x^2 + 8x + 13)(1 + x)'}{(1 + x)^2} | ||

+ | & = \frac{(8x + 8)(1 + x) - (4x^2 + 8x + 13)(1)}{(1 + x)^2} | ||

+ | & = \frac{4x^2 + 8x - 5}{(1 + x)^2} | ||

+ | f''(x) & = \frac{(4x^2 + 8x - 5)'(1 + x)^2 - (4x^2 + 8x - 5)((1 + x)^2)'}{(1 + x)^4} | ||

+ | \end{align*}</cmath> |

## Revision as of 20:57, 15 June 2020

## Problem 34

For the smallest value of is:

## Solution 1

To begin, lets denote the equation, as . Let's notice that:

After this simplification, we may notice that we may use calculus, or the AM-GM inequality to finish this problem because , which implies that both are greater than zero. Continuing with AM-GM:

Therefore, ,

## Solution 2 (Calculus)

Take the derivative of f(x) and f'(x) using the quotient rule.