Discuss why the equation f(x)=t has no real solutions if t

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.

(4x-1)(4x+1)=0

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.