Problem 68 - Calculus Single Variable

Question

Find a function \(f\) such that \(g \circ f=h\). \(g(x)=4 x+5, h(x)=8 x\)

Step-by-Step Solution

Answer

The function is \( f(x) = 2x - \frac{5}{4} \).

Step 1: Understand the Composition

We need a function \( f \) such that the composition of \( g \circ f \) equals \( h(x) \). This means substituting \( f(x) \) into \( g(x) \) should result in \( h(x) \). The equation becomes \( g(f(x)) = h(x) \).

Step 2: Set Up the Equation

From the composition requirement, substitute \( f(x) \) into \( g(x) \), resulting in the equation \( 4f(x) + 5 = 8x \). This originates from the expressions \( g(x) = 4x + 5 \) and \( h(x) = 8x \). Our goal is to solve for \( f(x) \).

Step 3: Solve for \( f(x) \)

Re-arrange the equation \( 4f(x) + 5 = 8x \) by isolating \( f(x) \). First, subtract 5 from both sides to get \( 4f(x) = 8x - 5 \). Then, divide both sides by 4 to solve for \( f(x) \): \[ f(x) = \frac{8x - 5}{4} \].

Step 4: Simplify \( f(x) \)

Simplify the expression for \( f(x) \):Divide \( 8x \) and \(-5\) by 4, leading to:\[ f(x) = 2x - \frac{5}{4} \]. This is the function that, when composed with \( g(x) \), gives \( h(x) \).

Step 5: Verify the Solution

Verify by checking if \( g(f(x)) \) truly equals \( h(x) \). Substitute \( f(x) \) into \( g(x) \):\( g(f(x)) = 4(2x - \frac{5}{4}) + 5 \).Calculate: \( g(f(x)) = 8x - 5 + 5 = 8x \), which equals \( h(x) \). Therefore, the solution is verified.

Key Concepts

Inverse Functions

In mathematics, an inverse function essentially "undoes" what the original function does. If you have a function represented by \( g(x) \) and its inverse represented by \( g^{-1}(x) \), applying \( g^{-1}(x) \) after \( g(x) \) will give you the starting value \( x \). It's like pressing "undo" on a calculator.
To find an inverse function, you need to swap the roles of \( x \) and \( y \) in the equation and then solve for the new value of \( y \). Keep in mind that not all functions have inverses. A function must be bijective (both injective and surjective) to have an inverse. This means it must have a one-to-one mapping between inputs and outputs.
Inverse functions are especially useful in solving equations and understanding function compositions, like in the exercise given where we had to ensure \( g(f(x)) \) leads back to \( h(x) \). Understanding inverse functions is essential for solving many algebraic problems effectively.

Function Transformation

Function transformation involves changing the appearance or the graph of a function, which is often achieved by manipulating the function's equation. It's like taking the basic form of a function and moving it around or stretching it.
There are several types of transformations, including:

• Translation (sliding the graph up, down, left, or right)
• Reflection (flipping the graph over a line, like the x-axis or y-axis)
• Stretching or compression (making the graph narrower or wider)

In the context of our problem, \( f(x) \) represents a kind of linear transformation, where we used algebraic manipulation to "transform" the function \( 4x + 5 \) so that, when \( f(x) \) is plugged into \( g(x) \), it transforms into \( h(x) = 8x \). Function transformation skills allow us to modify basic functions into more complex ones for specific needs.

Algebraic Manipulation

Algebraic manipulation involves using algebraic techniques to simplify or rearrange equations to isolate a variable, solve for a function, or simplify expressions. It's like rearranging a puzzle until each piece fits perfectly.
In the exercise, we used algebraic manipulation to find \( f(x) \) by solving the equation \( 4f(x) + 5 = 8x \). Here are the core steps we took:

• Subtracted 5 from both sides to eliminate the constant term on the left: \( 4f(x) = 8x - 5 \)
• Divided the entire equation by 4 to solve for \( f(x) \): \[ f(x) = \frac{8x - 5}{4} \]
• Simplified the expression for clarity: \( f(x) = 2x - \frac{5}{4} \)

These steps highlight how vital algebraic manipulation is in solving equations effectively. It’s about finding the simplest form of an equation while maintaining its equality. This is a fundamental skill in mathematics, underpinning much of algebra, calculus, and beyond.