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Find the value of \(x\) that maximizes

 

\(f(x) = \log (-20x + 12\sqrt{x})\)

 Jul 8, 2024
 #1
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The expression under the logarithm cannot be less than zero because the logarithm of a negative number is undefined.

 

Therefore, -20x + 12sqrt(x) must be greater than zero. This means:

 

12sqrt(x) > 20x

 

Square both sides of the inequality:

 

144x > 400x^2 - 480x + 144

 

Rearrange the inequality to get a quadratic expression:

 

400x^2 - 624x + 144 = 0

 

Divide both sides by 48 to simplify:

 

25x^2 - 39x + 9 = 0

 

This expression factors as:

(5x - 3)(5x - 3) = 0

 

Therefore, the possible solutions for x are:

x = 3/5

 

However, the original expression includes a square root. Since the square root of a negative number is undefined, we need to check if x = 3/5 makes -20x + 12sqrt(x) positive.

 

Plugging in x = 3/5:

-20(3/5) + 12sqrt(3/5) = -12 + 12sqrt(3/5)

 

Since 12sqrt(3/5) is positive, -12 + 12sqrt(3/5) is also positive. Therefore, x = 3/5 is a valid solution.

 

Answer: x = 3/5

 Jul 8, 2024

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