Let's break this down step by step, addressing each part of your question.

Given:

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

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a) Find $f'(x)$ (the derivative of $f(x)$)

To find the derivative, we apply the quotient rule: The quotient rule states that for two functions $u(x)$ and $v(x)$, the derivative of $\frac{u(x)}{v(x)}$ is: $\frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{v(x)u'(x) - u(x)v'(x)}{[v(x)]^2}$

Let:

• $u(x) = 2x + 1$
• $v(x) = x - 1$

Now, find the derivatives:

• $u'(x) = 2$
• $v'(x) = 1$

Using the quotient rule: $f'(x) = \frac{(x - 1)(2) - (2x + 1)(1)}{(x - 1)^2}$ Simplifying the numerator: $f'(x) = \frac{2x - 2 - 2x - 1}{(x - 1)^2} = \frac{-3}{(x - 1)^2}$

Thus, the derivative is: $f'(x) = \frac{-3}{(x - 1)^2}$

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b) Sketch the graph of $f(x)$ for $-3 \leq x \leq 5$

For sketching, let's analyze the behavior of the function.

• Domain: The function is undefined at $x = 1$ (division by zero), so there is a vertical asymptote at $x = 1$.
• Asymptotes:
  • Vertical asymptote: $x = 1$
  • Horizontal asymptote: As $x \to \infty$ or $x \to -\infty$, the degree of the numerator and denominator are both 1, so the horizontal asymptote is $y = 2$ (since the ratio of the leading coefficients is $\frac{2}{1} = 2$).
• Behavior near the asymptotes:
  • As $x \to 1^+$, $f(x) \to +\infty$
  • As $x \to 1^-$, $f(x) \to -\infty$
  • As $x \to \infty$, $f(x) \to 2$
  • As $x \to -\infty$, $f(x) \to 2$

You can plot the graph within the range $-3 \leq x \leq 5$, with the asymptotes in mind.

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c) Write down the equations of the asymptotes

From the analysis above:

• Vertical Asymptote: $x = 1$ (this is where the denominator equals zero).
• Horizontal Asymptote: $y = 2$ (since the function approaches 2 as $x \to \infty$ or $x \to -\infty$).

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d) Write down the x-intercept

The x-intercept occurs where $f(x) = 0$, i.e., where the numerator of the function equals zero.

Set the numerator equal to zero: $2x + 1 = 0$ Solving for $x$: $2x = -1 \quad \Rightarrow \quad x = -\frac{1}{2}$

Thus, the x-intercept is at $x = -\frac{1}{2}$.

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e) Solve $f(x) = f(x)$

The equation $f(x) = f(x)$ is always true for any $x$ in the domain of the function. It represents an identity, meaning that both sides are equal by definition. Therefore, there is no further solution or restriction beyond the domain.

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Summary of answers:

• a) $f'(x) = \frac{-3}{(x - 1)^2}$
• b) Graph sketch: Features a vertical asymptote at $x = 1$ and a horizontal asymptote at $y = 2$.
• c) Asymptotes:
  • Vertical: $x = 1$
  • Horizontal: $y = 2$
• d) x-intercept: $x = -\frac{1}{2}$
• e) $f(x) = f(x)$ is true for all values in the domain.

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Would you like any details on these points, or help with something else? Here are some related questions for further exploration:

1. How would the graph of $f(x)$ change if the numerator or denominator were modified?
2. What is the significance of the horizontal asymptote in rational functions?
3. How do you determine whether a rational function has any slant asymptotes?
4. Can we find the domain of any rational function in a similar way to this one?
5. How would the x-intercept behave if we changed the constants in the numerator?

Tip: When sketching rational functions, always consider the behavior near asymptotes and the end behavior (as $x \to \infty$) to get an accurate idea of the graph.