Let f(x) = 4x - x2, we have one solution to f(x) = 0 betwen [-1,0] since f(-1) < 0 and f(0) > 0.
Examine f'(x) = ln4 4x - 2x, and f'(x) > 0 when x < 0, so we have only one root when x < 0.
Now, let's examine x > 0. There seems no root for f(x) for x > 0. To prove that we again look at the derivative
f'(x) = ln4 4x - 2x, if we can prove that f'(x) > 0 for all x > 0, then we do not have any roots.
For simplicity, let's examine g(x) = 4x - 2x, let's look for real roots for this function.
Again examine the derivative: g'(x) = ln4 4x - 2, and this derivative a monotonically increasing function of x, and is zero at x=a,
where a = (ln2 - lnln4)/ln4
Since g'(x) is increasing and g'(a) = 0, g(x) increases for x > a. So, that's the point where g(x) is minimum over the real line. Let's look at that real value, see if it is positive or not.
g(a) = 4a - 2a = 2/ln4 - 2((ln2 - lnln4)/ln4) ? 0 (here ? denotes < or >, but we dont know yet)
1/ln4 ? (ln2 - lnln4)/ln4
1 ? ln2 - lnln4 = ln(2/ln4)
e ? 2/ln4 --> 2/ln4 is smaller than 2, and so smaller than e. So, ? is >.
Then, g(a) > 0, the minimum of g(x) occurs at x = a and it is greater than 0. So, we don't have root for 4x - 2x.
So, we cannot have root for ln4 4x - 2x since ln4 > 1. Then f'(x) > 0 for all x. Then f(x) is monotonically increasing and f(x) has only one real root which is between [-1,0].
