Problem 54

Question

For each pair of functions \(f(x)\) and \(g(x)\), find a. \(f(g(x))\) b. \(g(f(x))\) and c. \(f(f(x))\) $$ f(x)=x^{2}+1 ; g(x)=x-\sqrt{x} $$

Step-by-Step Solution

Verified
Answer
1. \(f(g(x)) = x^2 - 2x\sqrt{x} + x + 1\). 2. \(g(f(x)) = x^2 + 1 - \sqrt{x^2 + 1}\). 3. \(f(f(x)) = x^4 + 2x^2 + 2\).
1Step 1: Compute \(f(g(x))\)
First, substitute \(g(x)\) into \(f(x)\). The function \(g(x) = x - \sqrt{x}\) will replace every occurrence of \(x\) in \(f(x)\).Given \(f(x) = x^2 + 1\), substitute \(g(x)\) into \(f\): \[f(g(x)) = (g(x))^2 + 1 = (x - \sqrt{x})^2 + 1\] Now, expand the squared term:\[(x - \sqrt{x})^2 = x^2 - 2x\sqrt{x} + x\]Thus, \[f(g(x)) = x^2 - 2x\sqrt{x} + x + 1\].
2Step 2: Compute \(g(f(x))\)
Now, substitute \(f(x)\) into \(g(x)\). The function \(f(x) = x^2 + 1\) will replace every occurrence of \(x\) in \(g(x)\).Given \(g(x) = x - \sqrt{x}\), substitute \(f(x)\) into \(g\): \[g(f(x)) = (f(x)) - \sqrt{f(x)} = (x^2 + 1) - \sqrt{x^2 + 1}\].
3Step 3: Compute \(f(f(x))\)
Lastly, substitute \(f(x)\) into itself. The function \(f(x) = x^2 + 1\) will replace every occurrence of \(x\) in \(f(x)\).\[f(f(x)) = (f(x))^2 + 1 = (x^2 + 1)^2 + 1\]Expand the squared term:\[(x^2 + 1)^2 = x^4 + 2x^2 + 1\]Thus, \[f(f(x)) = x^4 + 2x^2 + 1 + 1 = x^4 + 2x^2 + 2\].

Key Concepts

Composite FunctionsNested FunctionsSubstituting Functions
Composite Functions
Composite functions involve taking one function and placing it inside another. This involves the operation of putting the output of one function into the input of another.

Think of it as a machine. You start with a value, which you send through the first machine (function) to get a new output. This output then goes into the next machine, and it alters the result once more.

In our example, for finding \( f(g(x)) \), we begin with function \( g(x) = x - \sqrt{x} \) and insert it into \( f(x) = x^2 + 1 \). So, wherever there's an \( x \) in \( f \), we replace it with \( g(x) \). The end result is, \( f(g(x)) = (x - \sqrt{x})^2 + 1 \).

When dealing with composite functions, it's important to carefully replace each variable correctly and handle any algebraic simplifications thereafter, as seen in our example.
Nested Functions
Nested functions are closely related to composite functions. When functions are nested, you find one function entirely within another function. This means the entire output of one function serves as the input for the entirety of another.

Consider the expression \( f(f(x)) \), where the function \( f \) is nested within itself. This means substituting \( f(x) = x^2 + 1 \) back into itself entirely. When computed, \( f(f(x)) \) gives you \((x^2 + 1)^2 + 1\).

It's like having layers or levels of functions. Each layer needs to be solved starting from the deepest one. In the case of \( f(f(x)) \), you must first resolve the innermost function completely before proceeding to the outer function.

Nested functions can become complex quickly with multiple layers, so always approach them step-by-step, simplifying each layer at a time.
Substituting Functions
Substitution in mathematics is the process of replacing one expression within an equation with another, equivalent one. When working with functions, substitution involves replacing one function with another function’s expression.

For instance, to find \( g(f(x)) \) from our original functions, you substitute \( f(x) = x^2 + 1 \) into \( g(x) = x - \sqrt{x} \). This becomes \( g(f(x)) = (x^2 + 1) - \sqrt{x^2 + 1} \).

To substitute successfully, plug the whole expression from one function everywhere \( x \) appears in the other.
  • Check each step to maintain accuracy.
  • Simplify the results wherever possible.
Substitution helps to see how changes in one function affect another, which is key in calculus and advanced mathematics.