1 Answer | Add Yours
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.
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].
We’ve answered 318,911 questions. We can answer yours, too.Ask a question