Chapter 6: Problem 72
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
Expert verified
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
01
- 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: \[ x = 2 \sqrt{y + 3} - 5 \]
02
- 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: \[ \frac{x + 5}{2} = \sqrt{y + 3} \]
03
- 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: \[ \frac{(x + 5)^2}{4} = y + 3 \]
04
- 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: \[ f^{-1}(x) = \frac{(x + 5)^2}{4} - 3 \]
05
- 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, \[ f^{-1}(f(x)) = x \]
06
- 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 \( f \): \[ R_{f^{-1}} = D_f = [-3, \infty) \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
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:
Let's break them down:
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)
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 \( 2\sqrt{x+3} - 5 \).
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:
Let's review the given function
\[ f(x) = 2 \sqrt{x + 3} - 5 \]:
The most basic square root function is \( f(x) = \sqrt{x} \). This function has:
- A domain of \[ [0, \infty) \]
- A range of \[ [0, \infty) \]
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.