Given the function f(x)=4^x + 5^x - 3^x - 6^x show that the equation f(a)=0 has the solutions 0 and 1.

We'll apply Lagrange's theorem to prove that the equation
f(a)=0, has the solutions 0 and 1.

We'll write the equation f(a) =
0.

4^a + 5^a - 3^a - 6^a = 0

We'll keep
to the left 4^a - 3^a and we'll move 5^a to the right:

4^a - 3^a =
6^a - 5^a

We'll create the function g(x) =
x^a.

We'll apply Lagrange's theorem to this function, over the range
[5 ; 6].

g(6)- g(5) = g'(c)(6-5)

6^a -
5^a = a*c^(a-1)

We'll apply Lagrange's theorem to this function,
over the range [3 ; 4].

g(4)- g(3) =
g'(d)(4-3)

4^a - 3^a = a*d^(a-1)

But
4^a - 3^a = 6^a - 5^a => a*c^(a-1) = a*d^(a-1)

We'll divide
by a:

c^(a-1) =d^(a-1)

(c/d)^(a-1) =
1

We'll create matching
bases:

(c/d)^(a-1) = (c/d)^0

This
equality proves either a-1 = 0, or c = d.

The last statement, c = d,
is impossible, since c is in the range (5 ; 6) and d is in the range (3 ;
4).

Only the identity a-1 = 0 holds.

a
= 1

The equation f(a) = 0 could have only 2 solutions
for a: {0 ; 1}.