Problem 22 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

\(f(x)=\frac{1}{2-x}\) and \(g(x)=x^{2}+1\). Find the following. Simplify your answers. If simplifying is dif cult, consult Appendix A: Algebra. (a) \(2 f(x+1)\) (b) \(f(2 x-2)\) (c) \(g(\sqrt{x}+1)\) (d) \(f(g(x))\) (e) \(g(f(x))\) (f) \(f(f(x))\) (g) \(g\left(\frac{1}{f(x)}\right)\) (h) \(\frac{g(x)}{f(x)}\)

Step-by-Step Solution

Answer

The simplified answers for the exercise are: (a) \(2 f(x+1) = \frac{2}{1-x} \)(b) \(f(2 x-2) = \frac{1}{6-2x} \)(c) \(g(\sqrt{x}+1) = x+2\sqrt{x}+2\)(d) \(f(g(x)) = \frac{1}{1-x^{2}}\)(e) \(g(f(x)) = (2-x)^{-2}+1\)(f) \(f(f(x)) = \frac{2-x}{2x-1}\)(g) \(g\left(\frac{1}{f(x)}\right)=(2x-1)^{2}+1\)(h) \(\frac{g(x)}{f(x)} = (x^{2}+1)*(2-x) \)

Step 1: Find \(2 f(x+1)\)

We begin by substituting \(x+1\) into function f(x). Hence, \( f(x+1)=\frac{1}{2-(x+1)}=\frac{1}{1-x}\). Multiply it by 2 to achieve the final result: \(2 f(x+1)=2* \frac{1}{1-x}=\frac{2}{1-x}\)

Step 2: Find \(f(2 x-2)\)

Next, we substitute \(2x-2\) into our function f(x). Thus, \(f(2 x-2)=\frac{1}{2-(2x-2)}=\frac{1}{2-2x+4}=\frac{1}{6-2x}\)

Step 3: Find \(g(\sqrt{x}+1)\)

We substitute \(\sqrt{x}+1\) into our function g(x):\(g(\sqrt{x}+1)=(\sqrt{x}+1)^{2}+1=x+2\sqrt{x}+2\)

Step 4: Find \(f(g(x))\)

We substitute g(x) into our function f(x). Hence, \(f(g(x))=\frac{1}{2-(x^{2}+1)}=\frac{1}{1-x^{2}}\)

Step 5: Find \(g(f(x))\)

Next, we substitute f(x) into our function g(x). This gives us \(g(f(x))=(\frac{1}{2-x})^{2}+1=(2-x)^{-2}+1\)

Step 6: Find \(f(f(x))\)

We substitute f(x) into itself, which yields: \(f(f(x))=\frac{1}{2-f(x)}=\frac{1}{2-\frac{1}{2-x}}=\frac{1}{\frac{2x-1}{2-x}}=\frac{2-x}{2x-1}\)

Step 7: Find \(g\left(\frac{1}{f(x)}\right))\)

Now let's substitute \(\frac{1}{f(x)}\) into the function g. We get: \(g\left(\frac{1}{f(x)}\right)=\left(\frac{1}{f(x)}\right)^{2}+1=\left(\frac{2-x}{1}\right)^{2}+1=(2x-1)^{2}+1\)

Step 8: Find \(\frac{g(x)}{f(x)}\)

Lastly we will look at the ratio of function g(x) to f(x) which yields: \(\frac{g(x)}{f(x)}=\frac{x^{2}+1}{\frac{1}{2-x}} = (x^{2}+1)*(2-x)\)

Key Concepts

Function Operations

Function operations involve adding, subtracting, multiplying, or dividing functions, similar to how we handle numbers. This is a great way to build more complex functions from simpler ones. In this exercise, we're looking at operations like multiplying and dividing functions.

Consider part (a) of the exercise, where the task is to find \(2f(x+1)\). This involves substitution, where you replace \(x\) in \(f(x)\) with \(x+1\). After substitution, you multiply by 2, simplifying to \(\frac{2}{1-x}\). This shows how operations can alter your function, creating something new from the existing rule.

Another part that demonstrates function operations is part (h) where the function \(\frac{g(x)}{f(x)}\) is computed. Here we divide \(g(x)\) by \(f(x)\), resulting in a more complex expression. This underscores the importance of careful substitution and arithmetic to combine functions. While it may seem tricky at first, following step-by-step transformations makes it manageable.

Composition of Functions

Composing functions, or plugging one function into another, is a powerful tool to create intricate relationships between variables. The notation \((f \circ g)(x)\) represents a composite function, meaning \(f(g(x))\). It's like nesting functions within one another.

In this exercise, step 4 asks us to find \(f(g(x))\). Here, \(g(x) = x^2 + 1\) is plugged into \(f(x) = \frac{1}{2-x}\). So, instead of \(x\), use \(x^2 + 1\) inside \(f\). This modification results in \(f(g(x)) = \frac{1}{1-x^2}\).

Another example is \(g(f(x))\) from step 5. Here, the function \(f(x) = \frac{1}{2-x}\) is substituted into \(g(x) = x^2 + 1\), resulting in \(g(f(x)) = (\frac{1}{2-x})^2 + 1\). Composition helps tailor functions to complex scenarios efficiently.

Simplification of Expressions

Simplifying expressions is a critical step for readability and further manipulation. When expressions are simplified, they become easier to interpret and work with.

Take step 6 of the exercise, which involves simplifying \(f(f(x))\). Initially, substituting \(f(x)\) into itself gives a rather unwieldy expression. However, by working through the algebra, we arrive at \(\frac{2-x}{2x-1}\), which is far simpler to handle. This shows the value of taking time to simplify complex results.

Another clear example is step 2, where \(f(2x-2) = \frac{1}{6-2x}\) is considerably easier to work with than its original form \(\frac{1}{2-(2x-2)}\), showing how simplification can facilitate further operations or calculus later on.

• Simplification reduces calculation errors.
• It makes results more understandable.
• It prepares functions for further analysis.

With practice, simplifying expressions becomes intuitive, greatly assisting both basic math tasks and advanced problem-solving.