Problem 83 - Algebra and Trigonometry

Question

In Exercise 65 of Section 3.6 you were asked to solve equations in which the unknowns were functions. Now that we know about inverses and the identity function (see Exercise \(82 ),\) we can use algebra to solve such equations. For instance, to solve \(f \circ g=h\) for the unknown function \(f\) we perform the following steps: $$\begin{aligned} f \circ g &=h \\ f \circ g \circ g^{-1} &=h \circ g^{-1} \\\ f \circ I &=h \circ g^{-1} \\ f &=h \circ g^{-1} \end{aligned}$$ So the solution is \(f=h \circ g^{-1} .\) Use this technique to solve the equation \(f \circ g=h\) for the indicated unknown function. (a) Solve for \(f,\) where \(g(x)=2 x+1\) and \(h(x)=4 x^{2}+4 x+7\) (b) Solve for \(g,\) where \(f(x)=3 x+5\) and \(h(x)=3 x^{2}+3 x+2\)

Step-by-Step Solution

Answer

(a) \( f(x) = x^2 + 6 \); (b) \( g(x) = x^2 + x - 1 \).

Step 1: Find the Inverse of g(x)

Given the function \( g(x) = 2x + 1 \), we need to find its inverse \( g^{-1}(x) \). To do this, set \( y = 2x + 1 \) and solve for \( x \):\[ y = 2x + 1 \]Subtract 1 from both sides:\[ y - 1 = 2x \]Divide by 2:\[ x = \frac{y - 1}{2} \]Therefore, \( g^{-1}(x) = \frac{x - 1}{2} \).

Step 2: Determine Function f Using Inverse of g(x)

Now, use the formula \( f = h \circ g^{-1} \) to find \( f \):\[ f(x) = h(g^{-1}(x)) \]Substitute \( g^{-1}(x) = \frac{x - 1}{2} \) into \( h(x) = 4x^2 + 4x + 7 \):\[ f(x) = h\left(\frac{x - 1}{2}\right) \]Calculate:\[ f(x) = 4\left(\frac{x - 1}{2}\right)^2 + 4\left(\frac{x - 1}{2}\right) + 7 \]Simplify each part to find \( f(x) \).

Step 3: Solve for f(x)

First expand the square:\[\left(\frac{x - 1}{2}\right)^2 = \frac{(x - 1)^2}{4} = \frac{x^2 - 2x + 1}{4}\]Substitute into the expression for \( f(x) \):\[ f(x) = 4\left(\frac{x^2 - 2x + 1}{4}\right) + 4\left(\frac{x - 1}{2}\right) + 7 \]\[ f(x) = (x^2 - 2x + 1) + 2(x - 1) + 7\] Simplify:\[ f(x) = x^2 - 2x + 1 + 2x - 2 + 7 \]\[ f(x) = x^2 + 6\]

Step 4: Set Up Expression for g(x)

To solve \( f \circ g = h \) for \( g \), we'll use the knowledge of inverses.Given \( f(x) = 3x + 5 \) and \( h(x) = 3x^2 + 3x + 2 \), we need to express \( g \) as \( f^{-1} \circ h \). First, find \( f^{-1}(x) \).

Step 5: Find the Inverse of f(x)

Set \( y = 3x + 5 \) and solve for \( x \):\[ y = 3x + 5 \]Subtract 5:\[ y - 5 = 3x \]Divide by 3:\[ x = \frac{y - 5}{3} \]Thus, \( f^{-1}(x) = \frac{x - 5}{3} \).

Step 6: Determine Function g Using Inverse of f(x)

Use the formula \( g = f^{-1} \circ h \):\[ g(x) = f^{-1}(h(x)) \]Substitute \( h(x) = 3x^2 + 3x + 2 \) into \( f^{-1}(x) = \frac{x - 5}{3} \):\[ g(x) = f^{-1}(3x^2 + 3x + 2) \]\[ g(x) = \frac{3x^2 + 3x + 2 - 5}{3} \]Simplify:\[ g(x) = \frac{3x^2 + 3x - 3}{3} \]\[ g(x) = x^2 + x - 1 \]

Step 7: Conclusion

The solved expressions for the unknown functions are:(a) To solve \( f \circ g = h \) for \( f \), we find \( f(x) = x^2 + 6 \).(b) To solve \( f \circ g = h \) for \( g \), we find \( g(x) = x^2 + x - 1 \).

Key Concepts

Function Composition

Function composition is a powerful mathematical concept where two functions are combined to form a new function. This is done by applying one function to the result of another. For instance, if we have two functions, \( f(x) \) and \( g(x) \), the composition is denoted as \( f \circ g(x) \). This means you first apply \( g \) to \( x \), then take the result and apply \( f \).
In practice, function composition helps us build complex transformations from simpler ones, making it a crucial tool in solving equations that involve unknown functions. By understanding how different functions interact through composition, we can manipulate and solve equations more effectively.

Algebraic Manipulation

Algebraic manipulation involves performing operations such as addition, subtraction, multiplication, and division to simplify expressions or solve equations. It’s like rearranging puzzle pieces to see the bigger picture. When it comes to solving equations with functions, algebraic manipulation allows us to isolate and hone in on what we're trying to find.
In our example, to solve \( f \circ g = h \), we use algebraic steps like solving for \( g^{-1} \) or rearranging terms to piece out the unknowns correctly. Each algebraic step requires careful calculations to ensure accuracy. This skill is crucial for handling more complex equation structures, especially ones involving inverses and identities.

Function Inverses

Function inverses essentially tell us how to reverse a function’s operation. If you have a function \( g(x) \), its inverse \( g^{-1}(x) \) will return you to the original input when applied to \( g(x) \). Think of it as the undo button in mathematics!
To find the inverse, set the function equation equal to \( y \) and solve it for \( x \). This process is exemplified in finding \( g^{-1}(x) \) where \( g(x) = 2x + 1 \). We set \( y = 2x + 1 \) and solve for \( x \), finding \( g^{-1}(x) = \frac{y - 1}{2} \). This stage transforms the equation allowing us to compose it with another function, such as \( h \) in the original step-by-step solution.

Solving Equations with Functions

Solving equations involving functions can seem daunting at first, but it becomes manageable by following a systematic approach utilizing function inverses and algebraic manipulation.
For example, to solve for \( f \) in \( f \circ g = h \), after determining \( g^{-1}(x) \), we substitute it into the composite function \( h(g^{-1}(x)) \). This substitution simplifies the function into a form we can work with, helping us isolate \( f(x) \). Similarly, solving for \( g \) utilizes a similar inverse function process with \( f(x) = 3x + 5 \), leading to calculating \( f^{-1} \).

• First, rearrange and solve for the outermost functions.
• Always find the inverse of the function involved to reverse operations.
• Utilize substitution to simplify composite functions.

By breaking down these tasks and following these structured steps, we can navigate equations that use function composition and inversion effectively.