+4 Find the value of \(x\) that maximizes
\(f(x) = \log (-20x + 12\sqrt{x})\)
+195 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
