Problem 21 - Essential Calculus Early Transcendentals

Question

Find a formula for the inverse of the function. $$ f(x)=1+\sqrt{2+3 x} $$

Step-by-Step Solution

Answer

The inverse formula is \( f^{-1}(x) = \frac{(x - 1)^2 - 2}{3} \).

Step 1: Replace f(x) with y

Start by replacing the function notation with a variable. This helps simplify the problem. Set \( y = 1 + \sqrt{2+3x} \).

Step 2: Isolate the square root

Subtract 1 from both sides to isolate the square root part of the equation. Thus, \( y - 1 = \sqrt{2 + 3x} \).

Step 3: Square both sides

To eliminate the square root, square both sides of the equation. This gives \( (y - 1)^2 = 2 + 3x \).

Step 4: Solve for x

To find the formula for the inverse, solve the equation for \( x \). Subtract 2 from both sides: \( (y - 1)^2 - 2 = 3x \), then divide by 3: \( x = \frac{(y - 1)^2 - 2}{3} \).

Step 5: Replace y with f^{-1}(x)

Since we set \( y = f(x) \), we replace \( y \) back with \( x \) to express the inverse function: \( f^{-1}(x) = \frac{(x - 1)^2 - 2}{3} \).

Key Concepts

Function Notation

Function notation is a shorthand way to express mathematical functions and their corresponding variables. In simple terms, when you see something like \( f(x) \), it is indicating that \( f \) is a function applied to the input variable \( x \). The result is a specific output, which depends on the input value of \( x \). Function notation helps us easily define and work with different functions without needing lengthy descriptions or explanations each time.

In our original exercise, \( f(x) = 1 + \, \sqrt{2 + 3x} \), this represents a unique function where for any value of \( x \), there is a specific rule to follow (i.e., add 2, multiply by 3, take the square root, then add 1).

• Clearly shows relationships between variables.
• Helps with identifying dependent and independent variables at a glance.
• Essential for defining inverses, like \( f^{-1}(x) \) in solving the exercise.

Understanding function notation is crucial when working with equations and calculus. It offers clarity and precision, foundational for deeper mathematical exploration.

Solving Equations

Solving equations is a fundamental skill in mathematics. It involves manipulating an equation to find the value(s) of unknown variables. Let's delve into how we approached solving for the inverse function in the original exercise.

The task required finding the full expression for \( x \) in terms of \( y \). Initially, we had: \[ y = 1 + \sqrt{2 + 3x} \] The first step was to isolate the square root term. We did this by subtracting 1 from both sides:\[ y - 1 = \sqrt{2 + 3x} \]This simple arithmetic manipulation starts unraveling the layers.

• Notice how isolating terms often means adding, subtracting, multiplying, or dividing across the equation.
• Every action on one side of the equation must be mirrored on the other.

After isolating the square root, the next phase was eliminating the square root itself by squaring both sides. Ultimately, these steps reveal how algebra is like peeling an onion, slowly revealing the solution underneath.

Square Root Elimination

Square root elimination is a process used to simplify an equation by removing the square root symbol. This is especially important in finding inverse functions, where the input expression is often under a square root.

In our example, we had \( y - 1 = \sqrt{2 + 3x} \). To eliminate the square root, both sides were squared, resulting in:\[ (y - 1)^2 = 2 + 3x \] Squaring is effective because it "undoes" the square root, allowing us to solve the equation more straightforwardly by moving terms around until \( x \) is isolated.

• This method only works when both sides being squared are guaranteed non-negative, to preserve equality.
• Be careful: squaring can introduce extraneous solutions, so it's vital to check after squaring.

By tackling the square root, the pathway to finding \( x \) became clearer, eventually leading to the inverse function's expression: \( f^{-1}(x) = \frac{(x - 1)^2 - 2}{3} \). Square root elimination is a key technique that unlocks complex algebraic manipulations.