/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 103 If \(f(x)=\frac{3 x+8}{2 x-A}\) ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 103

If \(f(x)=\frac{3 x+8}{2 x-A}\) and \(f(0)=2,\) what is the value of \(A ?\)

Short Answer

Expert verified
The value of \(A\) is \(-4\).

Step by step solution

01

- Substitute the Given Value in the Function

Start by substituting the given value of the function, which is when \(x = 0\), into the function \(f(x) = \frac{3x + 8}{2x - A}\). This will help in solving for \(A\).So, \(f(0) = \frac{3(0) + 8}{2(0) - A} = 2\)
02

- Simplify the Equation

Simplify the equation \(\frac{8}{-A} = 2\). This simplifies to \(8 = 2(-A)\).
03

- Solve for \(A\)

Divide both sides of the equation by 2 to isolate \(A\). \(8 = -2A \rightarrow -A = 4 \rightarrow A = -4\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Function Evaluation
Function evaluation is the process of finding the output of a function for a given input. In our exercise, we need to evaluate the function when the input is 0. The given rational function is:
\( f(x) = \frac{3x + 8}{2x - A} \)
To evaluate the function, we substitute 0 for x:
\( f(0) = \frac{3(0) + 8}{2(0) - A} = \frac{8}{-A} \)
This simplifies our equation. Properly substituting values makes function evaluation simple and straightforward.
Solving Equations
In solving equations, our goal is to isolate the variable we need to find. In the exercise, after evaluating the function at 0, we have:
\( f(0) = \frac{8}{-A} = 2 \)
To find the value of A, we need to solve this equation.
First, we remove the fraction by multiplying both sides of the equation by -A:
\[ 8 = 2(-A) \]
Next, we simplify the equation to isolate A:
\[ 8 = -2A \]
Dividing both sides by -2 gives:
\[ A = -4 \]
Solving equations involves systematic steps to isolate the variable. Remember to simplify step by step and do the same operations on both sides of the equation.
Finding Parameters in Functions
Finding parameters in functions is about determining unknown values within function definitions. In our exercise, we are given the rational function:
\( f(x) = \frac{3x + 8}{2x - A} \)
and the condition that \ f(0) = 2\. Our goal is to find A that satisfies this condition.
Starting with \[ f(0) = \frac{8}{-A} = 2 \]
and solving the equation helps us determine that:
\[ 8 = -2A \]
Thus,
\[ A = -4 \]
By understanding the structure of the function and using the given conditions, we can find unknown parameters. This process is essential for accurately defining functions and ensuring they meet specified criteria.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Write the standard form of the equation of a circle with center (3,-2) and radius \(r=\frac{\sqrt{6}}{2}\).

Find the average rate of change of \(h(x)=x^{2}-2 x+3\) (a) From -1 to 1 (b) From 0 to 2 (c) From 2 to 5

The shelf life of a perishable commodity varies inversely with the storage temperature. If the shelf life at \(10^{\circ} \mathrm{C}\) is 33 days, what is the shelf life at \(40^{\circ} \mathrm{C} ?\)

Find the domain of \(h(x)=\sqrt[4]{x+7}+7 x\).

In statistics, the standard normal density function is given by \(f(x)=\frac{1}{\sqrt{2 \pi}} \cdot \exp \left[-\frac{x^{2}}{2}\right]\) This function can be transformed to describe any general normal distribution with mean, \(\mu,\) and standard deviation, \(\sigma .\) A general normal density function is given by \(f(x)=\frac{1}{\sqrt{2 \pi} \cdot \sigma} \cdot \exp \left[-\frac{(x-\mu)^{2}}{2 \sigma^{2}}\right] .\) Describe the transformations needed to get from the graph of the standard normal function to the graph of a general normal function.
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Statistics

Read Explanation

Calculus

Read Explanation

Logic and Functions

Read Explanation

Discrete Mathematics

Read Explanation

Geometry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.