# Determine all values of t if 4x^2-lnx>t and t is positive?

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The first thing to do is to analyze the monotony of the function f(x). For this reason, we'll differentiate the function:

f'(x) = 8x - 1/x

We'll put f'(x) = 0.

8x - 1/x = 0

(8x^2 - 1)/x = 0

We notice that we have a difference of 2 squares at numerator:

(8x^2 - 1) = (2x*sqrt2 - 1)(2x*sqrt2 + 1)

f'(x) = 0 if and only if (2x*sqrt2 - 1)(2x*sqrt2 + 1)= 0.

2x*sqrt2 - 1= 0 => x = 1/2sqrt2 = sqrt2/4

2x*sqrt2 + 1 = 0

x = - sqrt2/4

The derivative is negative, over the range (0; sqrt2/4) and it is positive over the range (sqrt2/4; +infinite).

That means that the function is decreasing over the interval (0; sqrt2/4] and it is increasing over the range [sqrt2/4; +infinite).

So, the point f(sqrt2/4) is a local minimum point for the function.

Therefore, f(x)>=f(sqrt2/4)>=t

f(1/6) = 8/16 - ln(sqrt2/4) = 1/2 + ln sqrt2/4

So, a =< 1/2 + ln sqrt2/4

**All real values of "a" are located in the interval (-infinite ; 1/2 + ln** **sqrt2/4 ]. **