Answer:
The value of x is 0.09
Step-by-step explanation:
Here, the given function is,
[tex]f(x)=\log(-20x+12\sqrt{x})[/tex]
Differentiating with respect to x,
[tex]f'(x)=\frac{1}{-20x+12\sqrt{x}}(-20+\frac{12}{2\sqrt{x}})[/tex]
[tex]=\frac{-20+\frac{6}{\sqrt{x}}}{-20x+12\sqrt{x}}[/tex]
[tex]=\frac{-10+\frac{3}{\sqrt{x}}}{-10x+6\sqrt{x}}[/tex]
[tex]=\frac{-10\sqrt{x}+3}{2\sqrt{x}(-5x+3\sqrt{x})}[/tex]
Again differentiating with respect to x,
[tex]f''(x)=-\frac{(-10\sqrt{3}+3)(-15x+3\sqrt{3})}{4x\sqrt{x}(-5x+3\sqrt{x})}[/tex]
For maxima or minima,
[tex]f'(x) = 0[/tex]
[tex]\implies \frac{-10\sqrt{3}+3}{2\sqrt{x}(-5x+3\sqrt{3})}=0[/tex]
[tex]\implies x \approx 0.09[/tex]
At x = 0.09,
f''(x) = negative,
Hence, for x = 0.09, f(x) is maximum.
