1.2 Q-1

Question Statement

Given the following real-valued functions $f(x)$ and $g(x)$, find:

• (a) $f \circ g(x)$
• (b) $g \circ f(x)$
• (c) $f \circ f(x)$

Functions:

1. $f(x) = 2x + 1, g(x) = \frac{3}{x - 1}, x \neq 1$
2. $f(x) = \sqrt{x + 1}, g(x) = \frac{1}{x^2}, x \neq 0$
3. $f(x) = \frac{1}{\sqrt{x - 1}}, g(x) = (x^2 + 1)^2, x \neq 1$
4. $f(x) = 3x^4 - 2x^2, g(x) = \frac{2}{\sqrt{x}}, x \neq 0$

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Background and Explanation

To solve these, we need to understand function composition, which means evaluating one function inside another. For example:

• $f \circ g(x)$ means we first apply $g(x)$, and then apply $f$ to the result.
• Similarly, $g \circ f(x)$ means we apply $f(x)$ first and then $g$ to the result.

We will apply this concept to each of the given pairs of functions.

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Solution

(a) Find $f \circ g(x)$

1. For $f(x) = 2x + 1$ and $g(x) = \frac{3}{x - 1}$:

We need to compute $f \circ g(x) = f(g(x))$:

$$f \circ g(x) = f\left(\frac{3}{x-1}\right) = 2\left(\frac{3}{x-1}\right) + 1 = \frac{6}{x-1} + 1 = \frac{x + 5}{x - 1}$$

2. For $f(x) = \sqrt{x + 1}$ and $g(x) = \frac{1}{x^2}$:

We need to compute $f \circ g(x) = f(g(x))$:

$$f \circ g(x) = f\left(\frac{1}{x^2}\right) = \sqrt{\frac{1}{x^2} + 1} = \sqrt{\frac{x^2 + 1}{x^2}} = \frac{\sqrt{x^2 + 1}}{x}$$

3. For $f(x) = \frac{1}{\sqrt{x - 1}}$ and $g(x) = (x^2 + 1)^2$:

We need to compute $f \circ g(x) = f(g(x))$:

$$f \circ g(x) = f\left((x^2 + 1)^2\right) = \frac{1}{\sqrt{(x^2 + 1)^2 - 1}} = \frac{1}{\sqrt{x^4 + 2x^2}} = \frac{1}{x\sqrt{x^2 + 2}}$$

4. For $f(x) = 3x^4 - 2x^2$ and $g(x) = \frac{2}{\sqrt{x}}$:

We need to compute $f \circ g(x) = f(g(x))$:

$$f \circ g(x) = f\left(\frac{2}{\sqrt{x}}\right) = 3\left(\frac{2}{\sqrt{x}}\right)^4 - 2\left(\frac{2}{\sqrt{x}}\right)^2 = \frac{48}{x^2} - \frac{8}{x} = \frac{8(6 - x)}{x^2}$$

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(b) Find $g \circ f(x)$

1. For $f(x) = 2x + 1$ and $g(x) = \frac{3}{x - 1}$:

We need to compute $g \circ f(x) = g(f(x))$:

$$g \circ f(x) = g(2x + 1) = \frac{3}{(2x + 1) - 1} = \frac{3}{2x}$$

2. For $f(x) = \sqrt{x + 1}$ and $g(x) = \frac{1}{x^2}$:

We need to compute $g \circ f(x) = g(f(x))$:

$$g \circ f(x) = g\left(\sqrt{x + 1}\right) = \frac{1}{(\sqrt{x + 1})^2} = \frac{1}{x + 1}$$

3. For $f(x) = \frac{1}{\sqrt{x - 1}}$ and $g(x) = (x^2 + 1)^2$:

We need to compute $g \circ f(x) = g(f(x))$:

$$g \circ f(x) = g\left(\frac{1}{\sqrt{x - 1}}\right) = \left(\left(\frac{1}{\sqrt{x - 1}}\right)^2 + 1\right)^2 = \left(\frac{x}{x - 1}\right)^2$$

4. For $f(x) = 3x^4 - 2x^2$ and $g(x) = \frac{2}{\sqrt{x}}$:

We need to compute $g \circ f(x) = g(f(x))$:

$$g \circ f(x) = g\left(3x^4 - 2x^2\right) = \frac{2}{\sqrt{3x^4 - 2x^2}} = x \cdot \frac{2}{\sqrt{3x^2 - 2}}$$

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(c) Find $f \circ f(x)$

1. For $f(x) = 2x + 1$:

We need to compute $f \circ f(x) = f(f(x))$:

$$f \circ f(x) = f(2x + 1) = 2(2x + 1) + 1 = 4x + 3$$

2. For $f(x) = \sqrt{x + 1}$:

We need to compute $f \circ f(x) = f(f(x))$:

$$f \circ f(x) = f(\sqrt{x + 1}) = \sqrt{\sqrt{x + 1} + 1}$$

3. For $f(x) = \frac{1}{\sqrt{x - 1}}$:

We need to compute $f \circ f(x) = f(f(x))$:

$$f \circ f(x) = f\left(\frac{1}{\sqrt{x - 1}}\right) = \frac{1}{\sqrt{\frac{1}{\sqrt{x - 1}}}}$$

4. For $f(x) = 3x^4 - 2x^2$:

We need to compute $f \circ f(x) = f(f(x))$:

$$f \circ f(x) = f\left(3x^4 - 2x^2\right) = 3\left(3x^4 - 2x^2\right)^4 - 2\left(3x^4 - 2x^2\right)^2$$

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Key Formulas or Methods Used

• Function Composition: If $f(x)$ and $g(x)$ are two functions, then:
  • $f \circ g(x) = f(g(x))$
  • $g \circ f(x) = g(f(x))$
• Simplification: When simplifying, always apply operations inside the functions first, and then simplify the outer operations.

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Summary of Steps

1. For $f \circ g(x)$:

   • Plug $g(x)$ into $f(x)$.

2. For $g \circ f(x)$:

   • Plug $f(x)$ into $g(x)$.

3. For $f \circ f(x)$ and $g \circ g(x)$:

   • Apply the function to itself as if it’s another function.

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