Discuss why the equation f(x)=t has no real solutions if t<x0, where x0 is the critical value of f(x)=8x^2-lnx?
We'll determine first the critical points.
f'(x) = 16x - 1/x
To determine the critical points, we'll put f'(x)=0
16x - 1/x = 0
16x^2 - 1 = 0
Since it is a difference of 2 squares, we'll substitute it by the equivalent product.
Now, we'll set each factor as zero:
4x-1 = 0
x = 1/4
4x+1 = 0
x = -1/4
Since the function f(x) contains the term ln x, we'll impose the constraint that the domain of the function is (0 , +infinite).
Since the value x = -1/4 is not included in the domain (0,+infinite), we'll reject it.
The only critical point is x = 1/4.
We'll calculate the 2nd derivative to see if the extreme is a minimum or maximum point.
f"(x) = 16+ 1/x^2
It is obvious that f"(x)>0, so f(1/4) = 1/2 + ln 4 is a minimum point.
Now, we'll discuss the requested case: f(x)=t has no real solutions if t<x0 = 1/4
Since f(1/4) represents the minimum point of the function, then, it is no possible to have values of the given function smaller than the minimum point, so the equation f(x)=t has no real solutions.