Question

Show that $f : \mathbf{R}_+ \to [-5, \infty)$ given by $f(x) = 4x^2 + 4x - 5$ is both one-one and onto where $\mathbf{R}_+ = [0, \infty)$. Also, find $p \in \mathbf{R}_+$ such that $f(p) = 3$.

Hint:

To complete the square easily for quadratics like $4x^2 + 4x - 5$, rewrite it as $(2x+1)^2 - 1 - 5 = (2x+1)^2 - 6$. Setting this equal to $y$ makes isolating $x$ direct and bypasses the full quadratic formula setup entirely.

Answer 1

Concept:
• A function is one-one (injective) if $f(x_1) = f(x_2)$ implies $x_1 = x_2$ for all elements in the domain.
• A function is onto (surjective) if the range of the function is equal to its codomain, meaning for every element $y$ in the codomain, there exists at least one element $x$ in the domain such that $f(x) = y$.

Step 1: Prove that the function is One-One.
Let $x_1, x_2 \in \mathbf{R}_+ = [0, \infty)$ such that $f(x_1) = f(x_2)$: \[ 4x_1^2 + 4x_1 - 5 = 4x_2^2 + 4x_2 - 5 \] Cancel out $-5$ from both sides: \[ 4x_1^2 + 4x_1 = 4x_2^2 + 4x_2 \] Rearranging all terms to one side: \[ 4(x_1^2 - x_2^2) + 4(x_1 - x_2) = 0 \] Factoring using the difference of squares identity $a^2 - b^2 = (a-b)(a+b)$: \[ 4(x_1 - x_2)(x_1 + x_2) + 4(x_1 - x_2) = 0 \] Factor out the common term $4(x_1 - x_2)$: \[ 4(x_1 - x_2)(x_1 + x_2 + 1) = 0 \] This gives two possible cases:
• $x_1 - x_2 = 0 \implies x_1 = x_2$
• $x_1 + x_2 + 1 = 0 \implies x_1 + x_2 = -1$ Since the domain is restricted to non-negative real numbers ($\mathbf{R}_+ = [0, \infty)$), $x_1 \ge 0$ and $x_2 \ge 0$. Therefore, their sum plus one can never be zero ($x_1 + x_2 + 1 \ge 1$). Thus, the second case is impossible, leaving only $x_1 = x_2$. This confirms that $f(x)$ is one-one.

Step 2: Prove that the function is Onto.
Let $y \in [-5, \infty)$ be an arbitrary element in the codomain. Set $f(x) = y$ and solve for $x$: \[ 4x^2 + 4x - 5 = y \implies 4x^2 + 4x - (5 + y) = 0 \] This is a quadratic equation in terms of $x$. Applying the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: \[ x = \frac{-4 \pm \sqrt{4^2 - 4(4)(-(5 + y))}}{2(4)} = \frac{-4 \pm \sqrt{16 + 16(5 + y)}}{8} \] Factor out 16 inside the square root radical: \[ x = \frac{-4 \pm \sqrt{16(1 + 5 + y)}}{8} = \frac{-4 \pm 4\sqrt{6 + y}}{8} = \frac{-1 \pm \sqrt{6 + y}}{2} \] Since the domain requires $x \in [0, \infty)$, $x$ must be non-negative. Therefore, we must discard the negative root sign: \[ x = \frac{\sqrt{6 + y} - 1}{2} \] Now check if this value of $x$ remains valid inside the domain for all values of $y \in [-5, \infty)$: If $y \ge -5$, then $6 + y \ge 1 \implies \sqrt{6 + y} \ge 1 \implies \sqrt{6 + y} - 1 \ge 0$. Thus, $x \ge 0$, which belongs to $\mathbf{R}_+$. Since a valid preimage $x$ exists for every element in the codomain, the function is onto.

Step 3: Find the value of $p \in \mathbf{R_+$ such that $f(p) = 3$.}
Set up the equation: \[ 4p^2 + 4p - 5 = 3 \implies 4p^2 + 4p - 8 = 0 \] Dividing the entire equation by 4: \[ p^2 + p - 2 = 0 \] Factoring by splitting the middle term: \[ p^2 + 2p - p - 2 = 0 \implies p(p + 2) - 1(p + 2) = 0 \implies (p - 1)(p + 2) = 0 \] This yields $p = 1$ or $p = -2$. Since $p$ must belong to the non-negative domain ($\mathbf{R}_+$), we select $p = 1$.