Problem 11 - Calculus

Question

For the functions \(f\) and \(g\) find (a) \(f(g(1))\) (b) \(g(f(1))\) (c) \(f(g(x))\) (d) \(g(f(x))\) (e) \(f(t) g(t)\) $$f(x)=1 / x, g(x)=3 x+4$$

Step-by-Step Solution

Answer

(a) \(\frac{1}{7}\), (b) 7, (c) \(\frac{1}{3x + 4}\), (d) \(\frac{3}{x} + 4\), (e) \(\frac{3t + 4}{t}\)

Step 1: Evaluate g(1)

To find \(f(g(1))\), we first need to compute \(g(1)\). Substitute 1 for \(x\) in \(g(x) = 3x + 4\): \[g(1) = 3(1) + 4 = 3 + 4 = 7\].

Step 2: Evaluate f(g(1))

Now that we have \(g(1) = 7\), substitute this into \(f(x) = \frac{1}{x}\): \[f(g(1)) = f(7) = \frac{1}{7}\].

Step 3: Evaluate f(1)

To find \(g(f(1))\), first find \(f(1)\). Substitute 1 for \(x\) in \(f(x) = \frac{1}{x}\): \[f(1) = \frac{1}{1} = 1\].

Step 4: Evaluate g(f(1))

Using the result from Step 3, substitute 1 into \(g(x) = 3x + 4\): \[g(f(1)) = g(1) = 3(1) + 4 = 3 + 4 = 7\].

Step 5: Function Composition f(g(x))

Compute \(f(g(x))\) by substituting \(g(x) = 3x + 4\) into \(f(x) = \frac{1}{x}\): \[f(g(x)) = f(3x + 4) = \frac{1}{3x + 4}\].

Step 6: Function Composition g(f(x))

Compute \(g(f(x))\) by substituting \(f(x) = \frac{1}{x}\) into \(g(x) = 3x + 4\): \[g(f(x)) = g\left(\frac{1}{x}\right) = 3\left(\frac{1}{x}\right) + 4 = \frac{3}{x} + 4\].

Step 7: Evaluate f(t)

To evaluate \(f(t)g(t)\), first compute \(f(t)\) by substituting \(t\) into \(f(x) = \frac{1}{x}\): \[f(t) = \frac{1}{t}\].

Step 8: Evaluate g(t)

Substitute \(t\) into \(g(x) = 3x + 4\) to find \(g(t)\): \[g(t) = 3t + 4\].

Step 9: Multiply f(t) and g(t)

Using \(f(t) = \frac{1}{t}\) and \(g(t) = 3t + 4\) from Steps 7 and 8, find \(f(t)g(t)\): \[f(t)g(t) = \frac{1}{t}(3t + 4) = \frac{3t + 4}{t}\].

Key Concepts

Evaluation of Functions

When you encounter a problem involving evaluating functions, it's all about plugging numbers into given function formulas and simplifying the results. Consider two functions, say, a basic linear function like \(g(x) = 3x + 4\) and a reciprocal function like \(f(x) = \frac{1}{x}\). The evaluation process becomes straightforward when you substitute specific values, such as 1, into these functions.

• For \(g(1)\), substitute 1 for every \(x\) in the function: \(g(1) = 3(1) + 4 = 7\).
• Then use this result to evaluate another function, \(f\), where \(f(g(1)) = f(7) = \frac{1}{7}\).

It involves taking one step at a time: find the result from one function, then use that in plugging into another function. This systematic approach is key to accurate and efficient function evaluation.

Algebraic Expressions

Algebraic expressions form the foundation of many mathematical operations you will perform, such as function composition. In the provided exercise, you work with expressions like \(f(x) = \frac{1}{x}\) and \(g(x) = 3x + 4\). You often need to substitute these expressions into each other to form new expressions.

For example, when computing \(f(g(x))\), substitute the entire expression \(3x + 4\) into \(f(x)\). You get:

• \(f(g(x)) = f(3x + 4) = \frac{1}{3x + 4}\).

This process of substituting one algebraic expression into another forms the core of what function composition is about, creating a new function from old ones. Keep practicing these substitution steps to simplify and manage complex algebraic expressions with confidence.

Inverse Functions

Although the original exercise does not explicitly find inverse functions, understanding inverses is crucial when manipulating composed functions. An inverse function essentially "undoes" what the original function does. For a function \(f(x)\), its inverse \(f^{-1}(x)\) should satisfy \(f(f^{-1}(x)) = x\) and \(f^{-1}(f(x)) = x\).

Consider \(g(x) = 3x + 4\). If we wanted its inverse, we'd set \(y = 3x + 4\) and solve for \(x\):

• Rearrange: \(x = \frac{y - 4}{3}\).

Thus, \(g^{-1}(x) = \frac{x - 4}{3}\). Reverse compositional steps would then use these inverse functions, allowing the evaluation of how one function changes and returns to a former state. Understanding the concept of inverse functions gives deeper insight into solving equations and verifying the solutions you find via composition.