We will first proof that the function is one-to-one and then prove that it is onto.
One-to-one: Let x and y be real numbers in the interval \([\frac{1}{4},3]\) and suppose f(x)=f(y). By the definition of f, that means that \(\frac{4x-1}{2x+5} = \frac{4y-1}{2y+5}\).
Then \((4x-1)(2y+5)= (4y-1)(2x+5)\).
Expanding out both sides, we find that
\(8xy-2y+20x-5 = 8xy-2x+20y-5\).
So \(-2y+20x = -2x+20y\).
So 22x = 22y and thus x=y. So the function f is one-to-one.
Onto: Let y be a value arbitrarily chosen from the interval [0,1]. Consider the value \(x = \frac{5y+1}{4-2y}\). Since y is in [0,1], 5y+1 is between 1 and 6, and 4-2y is between 2 and 4. So \(\frac{5y+1}{4-2y}\) must lie between \(\frac{1}{4}\) and 3. Therefore, x lies in the domain of our function f. Furthermore, f(x) = y. Therefore, f is onto.
Since the function f is both one-to-one and onto, it is bijective.
Notice that the two subsections are labeled and the proof includes a brief introduction at the start and a brief wrap-up at the end. Such features make longer proofs easier to read.