Question

Problem: Find the value of $x$ that maximizes $$f(x) = \log (-20x + 12\sqrt{x}).$$ If there is no maximum value, write "NONE".

Answer 1

None~!! ;o lol.

have a wonderful dayy

Answer 2

Answer:

The value of x is 0.09

Step-by-step explanation:

Here, the given function is,

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

Differentiating with respect to x,

$f'(x)=\frac{1}{-20x+12\sqrt{x}}(-20+\frac{12}{2\sqrt{x}})$

$=\frac{-20+\frac{6}{\sqrt{x}}}{-20x+12\sqrt{x}}$

$=\frac{-10+\frac{3}{\sqrt{x}}}{-10x+6\sqrt{x}}$

$=\frac{-10\sqrt{x}+3}{2\sqrt{x}(-5x+3\sqrt{x})}$

Again differentiating with respect to x,

$f''(x)=-\frac{(-10\sqrt{3}+3)(-15x+3\sqrt{3})}{4x\sqrt{x}(-5x+3\sqrt{x})}$

For maxima or minima,

$f'(x) = 0$

$\implies \frac{-10\sqrt{3}+3}{2\sqrt{x}(-5x+3\sqrt{3})}=0$

$\implies x \approx 0.09$

At x = 0.09,

f''(x) = negative,

Hence, for x = 0.09, f(x) is maximum.