Problem 65 - Algebra and Trigonometry

Question

Solving an Equation for an Unknown Function suppose that $$\begin{aligned} g(x) &=2 x+1 \\ h(x) &=4 x^{2}+4 x+7 \end{aligned}$$ Find a function \(f\) such that \(f \circ g=h .\) (Think about what operations you would have to perform on the formula for \(g\) to end up with the formula for \(h . )\) Now suppose that $$\begin{array}{l}{f(x)=3 x+5} \\ {h(x)=3 x^{2}+3 x+2}\end{array}$$ Use the same sort of reasoning to find a function \(g\) such that \(f \circ g=h .\)

Step-by-Step Solution

Answer

For the first part, \( f(x) = x^2 + \frac{3}{2}x + \frac{15}{4} \). For the second part, \( g(x) = x^2 - 2 \).

Step 1: Understand the Problem

We need to find a function \( f(x) \) such that \( f(g(x)) = h(x) \). This means we want a composition of functions where \( g(x) \) is substituted into \( f(x) \) to produce \( h(x) \).

Step 2: Substitute g(x) into f(x)

Given \( g(x) = 2x + 1 \), we need to express \( h(x) \) in terms of \( g(x) \). Replacing \( x \) in \( h(x) = 4x^2 + 4x + 7 \) with \( g(x) = y \), we solve for \( x \) in terms of \( y \) and find \( x = \frac{y - 1}{2} \).

Step 3: Solve for f(x)

Replace \( x \) in \( h(x) = 4x^2 + 4x + 7 \) with \( g^{-1}(x) = \frac{x - 1}{2} \). Then, substitute back into \( f(g(x)) \) such that \( f(y) = 4\left(\frac{y - 1}{2}\right)^2 + 4\left(\frac{y - 1}{2}\right) + 7 \). Simplify this expression to find \( f(y) = y^2 + \frac{3}{2}y + \frac{15}{4} \).

Step 4: Verify the Solution

Ensure that substituting \( g(x) \) back into \( f(x) \) yields \( h(x) \) precisely. After simplifying \( f \left(2x + 1\right) \) using \( f(x) = x^2 + \frac{3}{2}x + \frac{15}{4} \), check if it matches \( 4x^2 + 4x + 7 \).

Step 5: Reverse Problem for Other Set of Functions

Now, attempt the second part where \( f(x) = 3x + 5 \) and \( h(x) = 3x^2 + 3x + 2 \). We need to find \( g(x) \).

Step 6: Assume g(x) and Solve

Assume a form for \( g(x) \) such that when substituted into \( f(x) \), it gives \( h(x) \). Since \( f(g(x)) \) must yield a quadratic \( h(x) \), try \( g(x) = ax^2 + bx + c \). By trial substitution and equating to \( 3x^2 + 3x + 2 \), find values for \( a \), \( b \), and \( c \) by solving the linear system of equations derived from the coefficients.

Step 7: Verification and Conclusion

Ensure that the chosen values of \( a \), \( b \), and \( c \) for \( g(x) \) achieve exactly \( h(x) = 3x^2 + 3x + 2 \) when substituted into \( f(x) \). Confirm that \( g(x) = x^2 - 2 \) is correct by checking \( f(x^2 - 2) \) yields \( 3x^2 + 3x + 2 \) upon simplification.

Key Concepts

Inverse Functions

Inverse functions help us reverse a process and go back to the original input from a known output. In simple terms, if a function takes you from Point A to Point B, the inverse function will bring you back from Point B to Point A. Mathematically, for a function - Let’s call it \( f(x) \) - Its inverse is noted as \( f^{-1}(x) \)

• If \( f(x) = y \), then \( f^{-1}(y) = x \)
• It is important that \( f(f^{-1}(x)) = x \) and \( f^{-1}(f(x)) = x \)

Let's take an example:- If \( g(x) = 2x + 1 \), to find its inverse, swap \( x \) and \( y \), solve for \( x \) which gives us \( x = \frac{y - 1}{2} \).This means the inverse function \( g^{-1}(x) = \frac{x - 1}{2} \).An inverse function will inter-relate with function composition, i.e., combining function and its inverse should return the original value. This exchange is crucial as seen when we reverse-engineer to find \( f \) or \( g \) from a set of equations.

Quadratic Functions

Quadratic functions are one of the essential types of polynomial functions you'll encounter. They have the general form:- \( h(x) = ax^2 + bx + c \)- Here, \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \).Quadratic functions graph as parabolas:- If \( a > 0 \), it opens upwards; if \( a < 0 \), it opens downwards.They can represent real-world problems, such as computing projectile motions or finding maximum profit levels.Let's see how they are involved in function composition:- In the exercise given, we have \( h(x) = 4x^2 + 4x + 7 \) being composed with \( g(x) = 2x + 1 \).- Breaking it down step-by-step, to get back \( f \) from \( f \circ g = h \), you set \( x = g^{-1}(x) \) and plug it into \( h(x) \).This operation binds quadratic concepts tightly with inverse and systematic solutions and enables us to identify function transformations through direct substitution.

System of Equations

A system of equations involves solving for more than one variable simultaneously, using several equations. It’s like assembling pieces of a puzzle to reveal a complete picture. In mathematics, these systems can be solved by various methods:

• Substitution: Solve one equation for a variable and substitute back.
• Elimination: Add or subtract equations to eliminate a variable.
• Graphical: Plot the equations and find their intersecting point.

In the context of function composition, let's consider solving for unknown functions.For instance:- Given \( f(x) = 3x + 5 \) and \( h(x) = 3x^2 + 3x + 2 \), we seek \( g(x) \).- Assume \( g(x) = ax^2 + bx + c \).- Substitute in \( f \circ g = h \) and compare coefficients to form a system of linear equations.By calculating correctly through matching coefficients, this approach helps to derive accurate formulas for creating and verifying composed functions. It is an essential tool, enabling us to decode complex systems and understand interdependencies in equations.