Question

The function \(f\) is one-to-one. (a) Find its inverse function \(f^{-1}\) and check your answer. (b) Find the domain and the range of \(f\) and \(f^{-1}\). $$ f(x)=2 \sqrt{x+3}-5 $$

Short answer

The inverse function is \( f^{-1}(x) = \frac{(x + 5)^2}{4} - 3 \). The domain and range of \( f \) are \([-3, \infty)\) and \([-5, \infty)\) respectively. The domain and range of \( f^{-1} \) are \([-5, \infty)\) and \([-3, \infty)\) respectively.

Step-by-step solution

- Replace and interchange variables

Start by rewriting the function as an equation using y: \[ y = 2 \sqrt{x + 3} - 5 \] Next, interchange the variables x and y to begin finding the inverse function: \[ x = 2 \sqrt{y + 3} - 5 \]

- Isolate the square root

Move the constant term on the right-hand side to the left-hand side:\[ x + 5 = 2 \sqrt{y + 3} \] Next, isolate the square root by dividing both sides by 2: \[ \frac{x + 5}{2} = \sqrt{y + 3} \]

- Eliminate the square root

Square both sides to eliminate the square root: \[ \left( \frac{x + 5}{2} \right)^2 = y + 3 \] Simplify the left-hand side: \[ \frac{(x + 5)^2}{4} = y + 3 \]

- Solve for y

Isolate y by subtracting 3 from both sides of the equation: \[ y = \frac{(x + 5)^2}{4} - 3 \] The inverse function is: \[ f^{-1}(x) = \frac{(x + 5)^2}{4} - 3 \]

- Verify the inverse function

To verify, combine f and its inverse and check if the result is the identity function: \[ f(f^{-1}(x)) = 2 \sqrt{\left( \frac{(x + 5)^2}{4} - 3 \right) + 3} - 5 \] Simplify inside the square root: \[ 2 \sqrt{ \frac{(x + 5)^2}{4} } - 5 = 2 \cdot \frac{|x + 5|}{2} - 5 = x + 5 - 5 = x \] Thus, \( f(f^{-1}(x)) = x \). Similarly, \[ f^{-1}(f(x)) = \frac{(2 \sqrt{x + 3} - 5 + 5)^2}{4} - 3 - 3 = x + 3 + 3 \] Simplifying this, \[ f^{-1}(f(x)) = x \]

- Find the domain and the range of f and f^{-1}

For the original function \( f \), determine the domain by setting the expression under the square root non-negative: \[ x + 3 \geq 0 \Rightarrow x \geq -3 \] So the domain of \( f \) is \[ D_f = [-3, \infty) \] The range of \( f \) can be obtained by finding the range of the transformation applied to the square root function. As \( y \) ranges from \( 0 \) to \( \infty \), \[ y = 2 \sqrt{x+3} - 5 \] When \( x = -3, y = 2 \sqrt{0} - 5 = -5 \)When \( x \to \infty, y \to \infty \)So the range of \( f \) is: \[ R_f = [-5, \infty) \]For the inverse, the roles switch. The domain of \( f^{-1} \) is the range of \( f \): \[ D_{f^{-1}} = R_f = [-5, \infty) \] The range of \( f^{-1} \) is the domain of \( f \): \[ R_{f^{-1}} = D_f = [-3, \infty) \]

Key concepts

Domain and Range

Understanding the domain and range of a function and its inverse is crucial.
The domain of a function represents all possible input values. For our function, this means looking at the expression under the square root sign and setting it to be non-negative:
\[ x + 3 \geq 0 \Rightarrow x \geq -3 \].
Therefore, the domain of \(f\) is:
\[ D_f = [-3, \infty) \].

Now, considering the range of the function, examine how the output values change.
As \(x\) increases from \(-3\) to \(\infty\), the output \(y = 2 \sqrt{x+3} - 5\) starts at \(-5\) and goes to \(\infty\). Thus, the range of \(f\) is:

\[ R_f = [-5, \infty) \].

Switching these for the inverse function, the domain of \(f^{-1}\) is the range of \(f\):
\[ D_{f^{-1}} = [-5, \infty) \]. Similarly, the range of \(f^{-1}\) is:
\[ R_{f^{-1}} = [-3, \infty) \].

By finding the domain and range properly, you can understand the complete behavior of the function.

Function Transformation

Function transformation involves translating, stretching, or reflecting graphs.
For our given function:
\[ f(x) = 2 \sqrt{x + 3} - 5 \],
\the transformations include:

• Horizontal translation left by 3 units (due to \(+3\) under the square root)
• Vertical stretch by a factor of 2 (since the square root function is multiplied by 2)
• Vertical translation down by 5 units (because of \(-5\) outside the square root)

Each transformation changes the graph's appearance in predictable ways.

Let's break them down:

• Start with the basic square root function \( \sqrt{x} \).
• Shift it left by 3 to get \( \sqrt{x+3} \).
• Stretch it vertically to form \( 2\sqrt{x+3} \).
• Finally, shift it down 5 units to find \( 2\sqrt{x+3} - 5 \).

Understanding these fundamental transformations helps you visualize and sketch the function accurately.

Square Root Function

Square root functions naturally have specific characteristics and behaviors.
The most basic square root function is \( f(x) = \sqrt{x} \). This function has:

• A domain of \[ [0, \infty) \]
• A range of \[ [0, \infty) \]

Additions and multiplications to this function, like in our problem, transform its graph.
Let's review the given function
\[ f(x) = 2 \sqrt{x + 3} - 5 \]:

• The domain changes based on the expression inside the square root.
  For example, \( x + 3 \) must be non-negative, setting limitations on \(x\).
• The range is influenced by any value outside the square root, such as the \(-5\).
• The multiplier in front (like the 2) changes the steepness of the curve, but the basic shape remains the same.

The insights gained from transforming the basic square root function help in analyzing more complex functions easily.