Inverse Functions — Question 6

Question 6

Let $f(x) = \sqrt{2x - 1}$ with domain $x \geq \dfrac{1}{2}$.

• Show that $f$ is one-to-one and has an inverse.

• Find the inverse function $f^{-1}(x)$.

• Determine the domain and range of $f^{-1}(x)$.

• Verify that $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$.

Question 6 - Solution

(a) One-to-One Check:

Since $f(x) = \sqrt{2x - 1}$ is a square root function with a linear expression inside and domain $x \geq \dfrac{1}{2}$, it is strictly increasing. Therefore, it passes the horizontal line test and is one-to-one.

(b) Find the inverse:

Let $y = \sqrt{2x - 1}$. Solve for $x$:

$$y^2 = 2x - 1 \Rightarrow 2x = y^2 + 1 \Rightarrow x = \frac{y^2 + 1}{2}$$

Now switch $x$ and $y$:

$$f^{-1}(x) = \frac{x^2 + 1}{2}$$

Answer: $\boxed{f^{-1}(x) = \dfrac{x^2 + 1}{2}}$

(c) Domain and Range of Inverse:

Since original domain is $x \geq \frac{1}{2}$, the range is:

$$f(x) = \sqrt{2x - 1} \geq 0 \Rightarrow \text{Range of } f = [0, \infty)$$

So: - Domain of $f^{-1}(x) = [0, \infty)$ - Range of $f^{-1}(x) = \left[ \frac{1}{2}, \infty \right)$

(d) Verifying composition:

$$f(f^{-1}(x)) = f\left( \frac{x^2 + 1}{2} \right) = \sqrt{2 \cdot \frac{x^2 + 1}{2} - 1} = \sqrt{x^2 + 1 - 1} = \sqrt{x^2} = x \quad (\text{since } x \geq 0)$$

$$f^{-1}(f(x)) = f^{-1}\left( \sqrt{2x - 1} \right) = \frac{( \sqrt{2x - 1} )^2 + 1}{2} = \frac{2x - 1 + 1}{2} = x$$

Both identities verified.