1Suppose that f is a function given as f(r) =2x + 1Simplify the expression f(x + h).%3Df(x + h) – f(x)Simplify the difference quotient,hf(x + h) - f(x)%3DhThe derivative of the function at a is the limit of the difference quotient as h approaches zero.f(x + h)- f(x)f'(x) = lim%3Dh

The given function is fx=12x+1. This implies that, fx+h=12x+h+1=12x+2h+1 Thus, fx+h=12x+2h+1.

Answered: 1 Suppose that f is a function given as…

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter4: Polynomial And Rational Functions

Section: Chapter Questions

Problem 15T

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Related questions

Question

1
Suppose that f is a function given as f(r) =
2x + 1
Simplify the expression f(x + h).
%3D
f(x + h) – f(x)
Simplify the difference quotient,
h
f(x + h) - f(x)
%3D
h
The derivative of the function at a is the limit of the difference quotient as h approaches zero.
f(x + h)- f(x)
f'(x) = lim
%3D
h

Expert Solution

Step 1

The given function is $f (x) = \frac{1}{2 x + 1}$ .

This implies that,

$\begin{array}{rcl} f (x + h) & = & \frac{1}{2 (x + h) + 1} \\ = & \frac{1}{2 x + 2 h + 1} \end{array}$

Thus, $\begin{array}{rcl} f (x + h) & = & \frac{1}{2 x + 2 h + 1} \end{array}$ .

Step 2

Obtain the difference quotient as follows.

$\begin{array}{rcl} \frac{f (x + h) - f (x)}{h} & = & \frac{\frac{1}{2 x + 2 h + 1} - \frac{1}{2 x + 1}}{h} \\ = & \frac{2 x + 1 - 2 x - 2 h - 1}{(2 x + 2 h + 1) (2 x + 1) h} \\ = & \frac{- 2 h}{(2 x + 2 h + 1) (2 x + 1) h} \\ = & - \frac{2}{(2 x + 2 h + 1) (2 x + 1)} \end{array}$