Problem 65 - A Graphical Approach to Precalculus with Limits

Question

For certain pairs of functions \(f\) and \(g .(f \circ g)(x)=x\) and \((g \circ f)(x)=x\). Show that this is true for the pairs in Exercises \(65-68\). $$f(x)=4 x+2, g(x)=\frac{1}{4}(x-2)$$

Step-by-Step Solution

Answer

The functions \(f(x) = 4x + 2\) and \(g(x) = \frac{1}{4}(x-2)\) are inverses.

Step 1: Understand the Problem

We have two functions: \( f(x) = 4x + 2 \) and \( g(x) = \frac{1}{4}(x-2) \). We need to verify that \((f \circ g)(x)=x\) and \((g \circ f)(x)=x\). This means we need to prove that \( f(g(x)) = x \) and \( g(f(x)) = x \).

Step 2: Calculate \(f(g(x))\)

In order to calculate \(f(g(x))\), substitute the expression for \(g(x)\) into \(f(x)\):\[f(g(x)) = f\left(\frac{1}{4}(x-2)\right)\]Substitute into function \(f\):\[f\left(\frac{1}{4}(x-2)\right) = 4 \left(\frac{1}{4}(x-2)\right) + 2\]Simplify:\[= x - 2 + 2 = x \]

Step 3: Verify \(f(g(x)) = x\)

We calculated that \(f(g(x)) = x\), which confirms that the first condition \((f \circ g)(x) = x\) is satisfied.

Step 4: Calculate \(g(f(x))\)

Now calculate \(g(f(x))\) by substituting \(f(x)\) into \(g(x)\):\[g(f(x)) = g(4x+2)\]Substitute into function \(g\):\[g(4x+2) = \frac{1}{4}((4x+2)-2)\]Simplify:\[= \frac{1}{4}(4x) = x\]

Step 5: Verify \(g(f(x)) = x\)

We calculated that \(g(f(x)) = x\), which confirms that the second condition \((g \circ f)(x) = x\) is satisfied.

Step 6: Conclusion

Since both \((f \circ g)(x) = x\) and \((g \circ f)(x) = x\) are satisfied, we have shown that the given functions \(f(x) = 4x + 2\) and \(g(x) = \frac{1}{4}(x-2)\) are indeed inverses of each other.

Key Concepts

Composition of Functions

Functions are like machines where you input something, and they give out a result. When we talk about the composition of functions, it's like feeding one machine into another. This means you take the output of one function and use it as the input for another. Suppose you have two functions, say \( f(x) \) and \( g(x) \). The composition \((f \circ g)(x)\) is the same as \(f(g(x))\). This simply means we first do \( g(x) \), and then feed its result into \( f(x) \), like a conveyor belt.

• Start with input \( x \).
• Process through \( g(x) \) to get a result.
• Feed that result into \( f \) for a final outcome.

Here, in the problem, the task was to check that \( (f \circ g)(x) = x \). This indicates that if you put \( x \) through \( g \) and then \( f \), you should end up with the original \( x \). It’s like solving a puzzle where the pieces fit perfectly to return to the start. Understanding composition helps you see how functions transform inputs step by step.

Function Verification

Verifying functions, especially in terms of composition, helps ensure that the functions work as anticipated. In the given exercise, we needed to check that both \((f \circ g)(x) = x\) and \((g \circ f)(x) = x\). Each equals \( x \) shows that both compositions revert the input to its initial value. When this happens, the functions are inverses of each other.Think of this as a way to double-check:

• Calculate \( f(g(x)) \) to ensure it simplifies back to \( x \).
• Calculate \( g(f(x)) \) for the same purpose.

Both results fitting perfectly confirms the condition that the composition fully "undoes" one another. If both checks hold true, it confirms that the functions are indeed inverse. Therefore, verification acts as both a checkpoint and a confirmation that our understanding and calculations align with mathematical truth.

Algebraic Manipulation

Algebraic manipulation allows us to rearrange and simplify expressions to understand them better. In verifying inverses, it aids in proving that one function can be reversed by another. Look at the problem where we simplify \( f(g(x)) \) and \( g(f(x)) \). For example:1. **Substituting values:** Begin with substituting \( g(x) \) into \( f(x) \). It simplifies to \( f(g(x)) = x - 2 + 2 \), which reduces to \( x \).2. **Simplification:** Continue by substituting \( f(x) \) into \( g(x) \), which gives \( g(f(x)) = \frac{1}{4}(4x) \), finally simplifying back to \( x \).Using these steps help shed light on how each piece plays its part in completing the picture. Algebraic manipulation is like fine-tuning details in a math problem to reveal the big picture - making complex expressions not just clear, but beautifully simple.