Problem 34 - Algebra 2

Question

Determine whether each pair of functions are inverse functions. $$ \begin{array}{l}{h(x)=5 x-7} \\ {g(x)=\frac{1}{5}(x+7)}\end{array} $$

Step-by-Step Solution

Answer

Yes, the functions are inverses of each other as both compositions equal \( x \).

Step 1: Understand Inverse Functions

Two functions \( f(x) \) and \( g(x) \) are inverses if \( f(g(x)) = x \) and \( g(f(x)) = x \) for every \( x \) in the domain. This means each function "undoes" the action of the other.

Step 2: Substitute \( g(x) \) into \( h(x) \)

To check if \( h(x) \) and \( g(x) \) are inverses, first calculate \( h(g(x)) \). Substitute \( g(x) = \frac{1}{5}(x+7) \) into \( h(x) \):\[ h(g(x)) = h\left(\frac{1}{5}(x+7)\right) = 5\left(\frac{1}{5}(x+7)\right) - 7. \]

Step 3: Simplify \( h(g(x)) \)

Simplify the expression obtained:\[ 5\left(\frac{1}{5}(x+7)\right) - 7 = (x + 7) - 7 = x. \]This shows that \( h(g(x)) = x \).

Step 4: Substitute \( h(x) \) into \( g(x) \)

Now, check the other direction by calculating \( g(h(x)) \). Substitute \( h(x) = 5x - 7 \) into \( g(x) \):\[ g(h(x)) = g(5x - 7) = \frac{1}{5}((5x - 7) + 7). \]

Step 5: Simplify \( g(h(x)) \)

Simplify the expression obtained:\[ \frac{1}{5}((5x - 7) + 7) = \frac{1}{5}(5x) = x. \]This shows that \( g(h(x)) = x \).

Step 6: Conclusion: Verify Both Conditions

Since we have shown both \( h(g(x)) = x \) and \( g(h(x)) = x \), the functions \( h(x) \) and \( g(x) \) are indeed inverse functions.

Key Concepts

Function Composition

Function composition plays a key role when verifying inverse functions. Imagine each function as a machine that transforms an input into an output.

• Function composition is like chaining together two machines, where the output of one becomes the input to another.
• For two functions, say \( f(x) \) and \( g(x) \), composition is denoted as \( f(g(x)) \).
• Here, \( g(x) \) acts first to transform an initial input, and then \( f(x) \) further processes the result.

When analyzing potential inverse functions, our goal is for these operations to cancel each other out:

• If \( f(g(x)) = x \) for every \( x \), then it's as if the machines combined just leave us with our original input.
• Both \( f(g(x)) \) and \( g(f(x)) \) should result in the same initial input value \( x \) for the functions to be inverse.

Algebraic Manipulation

Algebraic manipulation is the process of rearranging and simplifying expressions to reveal helpful insights or solutions.

• When verifying inverse functions, we need to substitute one function into the other and simplify the expression to see if we can get back to \( x \).
• Simplification often involves applying basic algebraic rules, like distributing terms and combining like terms.

For example, if you have an expression like \( 5\left(\frac{1}{5}(x+7)\right) - 7 \), the goal is to simplify it one step at a time:

• Distribute terms: \( 5\left(\frac{1}{5}(x+7)\right) = x+7 \).
• Then, simplify by combining terms: \( (x+7) - 7 = x \).

By performing these simplifications, we verify that the function compositions return us to \( x \), thus confirming the inverse relationship.

Domain of Functions

The domain of a function is the set of all possible inputs it can accept and is essential in understanding function composition, especially for inverse functions.

• When checking if two functions are inverses, ensure that both functions are defined across each other's range of output values.
• Practically, this means checking that the output of \( g(x) \) falls within the domain of \( h(x) \), and vice versa.

Considerations for domains become crucial due to the following reasons:

• If a function is not defined at some point, its inverse might not "undo" the operation as expected.
• This can lead to incomplete verification unless accounted for, as different function types have natural restrictions.

Thus, assessing domain constraints ensures that crucial portions of the function's behavior are not ignored while examining their inverses.