# Solute for a f(x)=>a if x>0, f(x)=18xto2-ln(x)?

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Notice that is f(x)>= a, then "a" must be a minimum point.

You need to calculate the minimum point of the function, hence, you should find the first derivative of the function.

`f'(x) = 2*18x - 1/x =gt f'(x) = (36x^2 - 1)/x`

Solving for x f'(x)=0, yields:

`(36x^2 - 1)/x = 0 =gt (36x^2 - 1) = 0`

You need to write the difference of squares such that:

`(36x^2 - 1) = (6x - 1)(6x + 1) =gt (6x - 1)(6x + 1) = 0`

`6x - 1 = 0 =gt 6x = 1 =gt x = 1/6`

`6x + 1 = 0 =gt x = -1/6`

The function increases over `(-oo ;-1/6]` and `[1/6; oo)` and it decreases over `(-1/6 ; 1/6).`

The minimum point is at `x = 1/6`

Hence `f(x)gt= a`

`f(x) gt= f(1/6) = 18/36 - ln (1/6)`

`a<= 1/2 - (ln 1 - ln 6) =gt a<= 1/2 - 0 + ln 6`

`` `a<= 1/2 + ln 6`

**The solution to the inequality `f(x)gt= a` `is ` the interval `[1/2 + ln 6 ; oo).` **