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Chapter 1: Problem 4

If \(f(x)=\frac{x-1}{x+1}\), then \(f(2 x)\) in terms of \(f(x)\) is (a) \(\frac{f(x)+1}{f(x)+3}\) (b) \(\frac{3 f(x)+1}{f(x)+3}\) (c) \(\frac{f(x)+3}{f(x)+1}\) (d) \(\frac{f(x)+3}{3 f(x)+1}\)

Short Answer

Expert verified
The answer is (d) \(\frac{f(x)+3}{3f(x)+1}\).

Step by step solution

01

Understand the function given

The function given in the problem is \(f(x) = \frac{x-1}{x+1}\). This is the function whose transformation we need to explore.
02

Substitute 2x into the function

Substituting \(2x\) in the place of \(x\) in the given function, we get \(f(2x) = \frac{2x-1}{2x+1}\). This is the result for substituting \(2x\) into \(f(x)\).
03

Express f(2x) in terms of f(x)

We know \(f(x) = \frac{x-1}{x+1}\). We'll manipulate \(f(2x) = \frac{2x-1}{2x+1}\) to express it in terms of \(f(x)\).
04

Use f(x) expression

Given \(f(x) = \frac{x-1}{x+1} \Rightarrow f(x) + 1 = \frac{2x}{x+1}\). We use these expressions to find a relationship.
05

Relate manipulated equations

By using the identities and manipulation, examine relation derivation: \(f(2x) = \frac{2x-1}{2x+1} = \frac{f(x)+3}{3f(x)+1}\). See how transformations maintain equivalence between both sides through algebraic manipulation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Algebraic Manipulation
Algebraic manipulation is a fundamental tool in calculus that involves rearranging expressions to achieve a desired form. In this exercise, we started with a given function, \( f(x) = \frac{x-1}{x+1} \), and substituted \( 2x \) to find \( f(2x) = \frac{2x-1}{2x+1} \). To express \( f(2x) \) in terms of \( f(x) \), algebraic manipulation is used:
  • We observe that \( f(x) + 1 = \frac{2x}{x+1} \).
  • This step prepares us to find connections between \( f(x) \) and \( f(2x) \).
  • Through skilled algebraic manipulation, equations are adjusted, expanded, and arranged to unveil the relationship \( f(2x) = \frac{f(x)+3}{3f(x)+1} \).
This arrangement allows us to interpret \( f(2x) \) by leveraging the form of \( f(x) \). As you encounter similar problems, remember:
  • Manipulate neatly to avoid errors.
  • Look for simplifications that transform complicated expressions into more familiar forms.
  • Verify each step to ensure the equation remains balanced and equivalent.
Algebraic manipulation not only simplifies the problem-solving process but also deepens your understanding of functional relationships in calculus.
Function Transformation
Function transformation is a crucial concept in calculus involving modifications to the base form of functions. For the exercise in question, transforming \( f(x) \) to \( f(2x) \) is a prime example.
  • Start with the base function \( f(x) = \frac{x-1}{x+1} \).
  • By replacing \( x \) with \( 2x \), a horizontal transformation occurs, compressing the graph along the x-axis.
  • The transformed function, \( f(2x) = \frac{2x-1}{2x+1} \), maintains the core characteristics of \( f(x) \) but at twice the rate on the x-axis.
Understanding these transformations helps you:
  • Visualize how the function's graph changes with different modifications.
  • Predict the behavior of the function under various alterations.
  • Grasp the broader implications of functional transformation in calculus workflows.
These transformations can seem straightforward but are powerful in reshaping entire problem scenarios in calculus.
Substitution Method
The substitution method is a strategic approach for simplifying functions and expressions by replacing variables to reveal new insights. In this exercise:
  • We substitute \( 2x \) into \( f(x) \), resulting in \( f(2x) = \frac{2x-1}{2x+1} \).
  • This direct substitution helps in assessing how the function reacts to changes in its variable.
  • Additionally, expressing \( f(2x) \) in terms of \( f(x) \) requires another subtle substitution to showcase \( f(x) + 1 = \frac{2x}{x+1} \).
The substitution method is not just about plugging in numbers or expressions but also about:
  • Providing deeper insights into the function's behavior and structure.
  • Aiding in solving complex problems by reducing them into more manageable tasks.
  • Creating a bridge between different forms of a function to reveal hidden relationships.
Mastering substitution enhances problem-solving agility and opens avenues for exploring intricate calculus problems efficiently.

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Most popular questions from this chapter

If \(f(x)=x^{2}-3 x+1\) and \(f(2 \alpha)=2 f(\alpha)\), then \(\alpha\) is equal to (a) \(\frac{1}{\sqrt{2}}\) (b) \(-\frac{1}{\sqrt{2}}\) (c) \(\frac{1}{\sqrt{2}}\) or \(-\frac{1}{\sqrt{2}}\) (d) None of these

If \(e^{x}=y+\sqrt{1+y^{2}}\), then \(y\) is equal to (a) \(e^{x}+e^{-x}\) (b) \(e^{x}-e^{-x}\) (c) \(\frac{1}{2}\left(e^{x}-e^{-x}\right)\) (d) \(\frac{1}{2}\left(e^{x}+e^{-x}\right)\)

If \(e^{f(x)}=\frac{10+x}{10-x}, x \in(-10,10)\) and \(f(x)\) \(=k f\left(\frac{200 x}{100+x^{2}}\right)\), then \(k\) is equal to (a) \(0.8\) (b) \(0.7\) (c) \(0.6\) (d) \(0.5\)

If \(f: R \rightarrow R, f(x+y)=f(x)+f(y), \forall x, y \in R\) and \(f(1)=7\), then \(\sum_{r=1}^{n} f(r)\) is equal to (a) \(\frac{7 n(n+1)}{2}\) (b) \(\frac{7 n}{2}\) (c) \(\frac{7(n+1)}{2}\) (d) \(7 n(n+1)\)

If a function \(F\) is such that \(F(0)=2, F(1)=3\), \(F(n+2)=2 F(n)-F(n+1)\) for \(n \geq 0\), then \(F(5)\) is equal to (a) \(-7\) (b) \(-3\) (c) 7 (d) 13
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