Proving a quadratic function is bijective in given range

Question

Prove that $f: [0,\infty[ \to [-5,\infty[$ defined as $f(x) = 4x^2+4x-5$ is bijective. I can prove it graphically but not algebraically.

Answer 1

It is surjective: given some $y \in [-5,\infty)$, we can solve the equation $f(x) = y$ by $$ 4x^2 + 4x-5 = y \implies x = \frac{\pm\sqrt{y+6}-1}{2}. $$ Because $y$ is bigger than $-5$, this is well-defined so indeed $f$ is surjective. Furthermore, the negative branch of the square root, i.e. the solution $x=\frac{-\sqrt{y+6}-1}{2}$ will always be negative. So, in the given domain $[0,\infty)$, the solution for $x$ is unique, proving that $f$ is injective.

Answer 2

You can do this: \begin{equation} f(x_1)=f(x_2) \implies 4x^2_1+4x_1-5=4x^2_2+4x_2-5 \implies x_1^2-x_2^2+x_1-x_2=0 \implies\\ (x_1-x_2)(x_1+x_2)+(x_1-x_2)=0 \implies (x_1-x_2)(x_1+x_2+1)=0 \end{equation} Since $x_1,x_2 \geq 0$, it means $x_1+x_2+1 > 0$. This implies that $x_1-x_2=0 \implies x_1=x_2$, meaning the function is injective.

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Surjection: \begin{equation} f(x)=y \implies 4x^2+4x-5=y \end{equation} Ordinary quadratic equation. Solve it and prove that x is real and positive for $y\geq -5$ (the part under the root has to be non-negative). Note: I have not done the calculation myself.