Definition of deriviative, factoring problem

Click For Summary
SUMMARY

The discussion focuses on using the definition of the derivative to find f'(c) for the function f(x) = x^3 - 4x^2 + x + 8 at c = 1. The limit definition of the derivative is applied, leading to an indeterminate form when substituting x = 1. Participants suggest factoring the numerator, specifically identifying (x - 1) as a factor, and recommend using synthetic division to simplify the expression efficiently.

PREREQUISITES
  • Understanding of limits and continuity in calculus
  • Familiarity with the definition of the derivative
  • Knowledge of polynomial factoring techniques
  • Experience with synthetic division for polynomials
NEXT STEPS
  • Practice finding derivatives using the limit definition with various polynomial functions
  • Learn polynomial factoring methods, focusing on identifying roots
  • Study synthetic division and its applications in calculus
  • Explore the relationship between derivatives and graph behavior
USEFUL FOR

Students studying calculus, particularly those learning about derivatives and polynomial functions, as well as educators looking for effective teaching strategies in these topics.

gabby989062
Messages
20
Reaction score
0

Homework Statement


use definition of derivitiave to find f'(c) EXACTLY
x3 - 4x2 + x + 8, c=1

Homework Equations


lim x->c f(x)-f(c) / x-c


The Attempt at a Solution



I got as far as (x3 - 4x2 +x + 2) - 6 / x-1

Plugging in 1 for x gives me divided by zero. The teacher's example, she factored an x-1 in the numerator and canceled it with the denominator, but she did an easy example that only had x2, and I have x3. How do I find an x-1 in the top?
:smile::zzz::biggrin:
 
Last edited:
Physics news on Phys.org
Simplify the numerator, and try factoring first two terms, and then the next two terms.
 
By the way, it is much easier to read what you really mean if you use "^" to indicate powers: x^3- 4x^2+ x+ 2, rather than x3-4x2+x+2.

x^3- 4x^2+ x+ 2- 6= x^3- 4x^2+ x- 4= (x-1)(?).

Knowing that one factor must be x- 1 should make that easy- just divide x^3- 4x^2+ x- 4 by x-1. If you use "synthetic division" it's even easier.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 9 ·
Replies
9
Views
4K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 17 ·
Replies
17
Views
2K
Replies
9
Views
3K