Problem 11 - Calculus with Concepts in Calculus

Question

Let \(f(x)=\frac{x-1}{x^{2}+1}\) and \(g(x)=x^{1 / 4}\). Find the specified values. $$ (f g)(9) $$

Step-by-Step Solution

Answer

\((fg)(9) \approx 0.169\).

Step 1: Understand the Composition of Functions

To find \((fg)(x)\), we need to first recognize that this represents the product of the functions \(f(x)\) and \(g(x)\). Thus, \((fg)(x) = f(x) \cdot g(x)\). This means for \((fg)(9)\), we need to evaluate both functions \(f(x)\) and \(g(x)\) at \(x = 9\) and multiply the results.

Step 2: Evaluate f(x) at x=9

The function \(f(x)\) is given by \(f(x) = \frac{x-1}{x^{2}+1}\). Substitute \(x = 9\) into this function:\[f(9) = \frac{9-1}{9^2+1} = \frac{8}{81+1} = \frac{8}{82} = \frac{4}{41}.\]

Step 3: Evaluate g(x) at x=9

The function \(g(x)\) is given by \(g(x) = x^{1/4}\). Substitute \(x = 9\) into this function:\[g(9) = 9^{1/4}.\]The fourth root of 9 is the square root of the square root of 9. Calculate it as:\[9^{1/4} = \sqrt{\sqrt{9}} = \sqrt{3} = \approx 1.732.\]

Step 4: Calculate (fg)(9)

Now that we have \(f(9) = \frac{4}{41}\) and \(g(9) \approx 1.732\), we calculate their product:\[(fg)(9) = f(9) \cdot g(9) = \frac{4}{41} \cdot 1.732.\] This results in approximately:\[(fg)(9) \approx \frac{6.928}{41} \approx 0.169.\]

Key Concepts

Product of Functions

When you come across the notation \((fg)(x)\), it indicates the product of two functions, \(f(x)\) and \(g(x)\). Instead of merely adding or combining the functions, this involves multiplying the output of each function for a given input \(x\). For example, if \(f(x)\) and \(g(x)\) represent any two mathematical processes or operations, then \((fg)(x)\) is what you get by executing both operations on the same \(x\) and multiplying the results.
The process can be summarized in a few simple steps:

• Find the expression for each function, given the value of \(x\).
• Evaluate each function separately at \(x\).
• Take the product of these evaluated results.

Returning to our exercise, we first identified \(f(x)\) and \(g(x)\). The direct multiplication of their evaluations, as shown by \((fg)(9) = f(9) \cdot g(9)\), is the essence of function composition through multiplication. This method helps us in situations where we combine distinct processes to study their combined effect with a specific value.

Evaluating Functions

Evaluating a function involves substituting a specific value for the variable \(x\), and then performing arithmetic to simplify and find the result. For the problem at hand, we needed to determine the outputs of \(f(x)\) and \(g(x)\) at \(x = 9\).

To evaluate \(f(x) = \frac{x-1}{x^2+1}\) at \(x = 9\):

• Substitute 9 for every \(x\) in \(f(x)\).
• Carry out the arithmetic: calculate \(9-1\) and \(9^2 + 1\).
• Simplify to get \(f(9) = \frac{4}{41}\).

Similarly, for \(g(x) = x^{1/4}\) at the same \(x=9\):

• Substitute 9 for \(x\).
• Calculate the fourth root of 9 using \(9^{1/4}\), which simplifies to \(\sqrt{3}\).
• This gives us approximately \(g(9) = 1.732\).

Evaluating functions correctly is crucial to finding accurate solutions in more complex compositions.

Algebraic Manipulation

Algebraic manipulation is the skill of rearranging and simplifying expressions and equations. It plays an essential role when dealing with functions, especially during evaluation and multiplication.

• Simplifying Expressions: For \(f(9)\), you first rewrite the expression by making the necessary substitutions and then simplify by carrying out the arithmetic operations \( \frac{9-1}{9^2+1} \).
• Approximation: When dealing with expressions like \(9^{1/4}\), estimating values, such as approximating \(\sqrt{3} \approx 1.732\), becomes useful for practical purposes.
• Multiplication of Results: The knowledge of exact and estimated values helps in calculating \((fg)(9)\), resulting in \((fg)(9) \approx \frac{6.928}{41} \approx 0.169\).

These steps provide insight into transforming expressions to simpler forms, which is a highly valuable skill for tackling more complex problems and ensuring clarity in computation. This manipulation allows us to solve efficiently and confidently.