/*! 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 14 Find the derivative by the limit... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the derivative by the limit process. \(f(x)=9-\frac{1}{2} x\)

Short Answer

Expert verified
The derivative of the function \(f(x) = 9 - \frac{1}{2}x\) is \(f'(x) = -\frac{1}{2}\).

Step by step solution

01

Understand the problem

The problem is to find the derivative of the function \(f(x) = 9 - \frac{1}{2}x\). This demands us to apply the limit definition of the derivative.
02

Apply the limit definition of the derivative

To do so, find the limit as \(h\) approaches zero of: \[\frac{f(x+h)-f(x)}{h}\] Where \(f(x+h)\) involves replacing every instance of \(x\) in the original function with \((x+h)\), and \(f(x)\) is the original function.
03

Performing the calculations

\[f(x + h) = 9 - \frac{1}{2}(x + h)\] Next, replace \(f(x + h)\) and \(f(x)\) in the definition with obtained and original function respectively: \[\frac{f(x + h) - f(x)}{h} = \frac{(9 - \frac{1}{2}(x + h)) - (9 - \frac{1}{2}x)}{h}\] Simplify the above equation to find the limit as \(h\) approaches zero.
04

Simplification

After simplification, the equation becomes: \[\lim_{h\to0} -\frac{1}{2}\] Note: This simplifies directly because \(-\frac{1}{2}\) does not depend on \(h\). This concludes that \(f'(x) = -\frac{1}{2}\).

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 91Ó°ÊÓ!

One App. One Place for Learning.

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

Get started for free

Study anywhere. Anytime. Across all devices.