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how to simplify the difference quotient
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Question:Given the function f(x)=42xx^2
Evaluate the difference quotient for the given function and simplify the answer. [f(1+h)f(1)]\h
I've tried to work this problem several times on an online homework website but it keeps saying my answer is incorrect, so I don't know what I'm doing wrong. Can someone help work this out for me?
Answers:f(1+h) = 4  2(1+h)  (1+h) = 4  2  2h  1  2h  h = 4  2  1  2h  2h  h = 1  4h  h f(1) = 4  2  1 = 1 f(1+h)  f(1) = 4h  h [f(1+h)  f(1)] / h = (4h + h ) / h = h(4+h) / h For h 0 we may cancel the factor of h from numerator and denominator, leaving [f(1+h)  f(1)] / h = (4+h)
Answers:f(1+h) = 4  2(1+h)  (1+h) = 4  2  2h  1  2h  h = 4  2  1  2h  2h  h = 1  4h  h f(1) = 4  2  1 = 1 f(1+h)  f(1) = 4h  h [f(1+h)  f(1)] / h = (4h + h ) / h = h(4+h) / h For h 0 we may cancel the factor of h from numerator and denominator, leaving [f(1+h)  f(1)] / h = (4+h)
Question:The difference quotient is the basis of beginning to study a first Calculus course. Introductory Calculus is heavily based on Algebra. The difference quotient of a function f(x) is defined as f(x+h)f(x) / h
You can find the difference quotient in steps, if desired
1.Find the expression for f(x + h).
2.Find the expression for f(x + h)  f(x).
3.Find the expression for the difference quotient.
4.Simplify, if possible.
Find the difference quotient for f(x)= (x+1)^3.
Answers:1. f(x+h) = ((x+h) + 1)^3 2. f(x+h)  f(x) = ((x+h) + 1)^3 (x+1)^3 3. (f(x+h)  f(x))/h = (((x+h) + 1)^3 (x+1)^3)/h 4. = (3x^2h +3xh^2 +6xh +3h +3h^2 +h^3)/h = 3x^2 +3xh +6x +3 +3h + h^2
Answers:1. f(x+h) = ((x+h) + 1)^3 2. f(x+h)  f(x) = ((x+h) + 1)^3 (x+1)^3 3. (f(x+h)  f(x))/h = (((x+h) + 1)^3 (x+1)^3)/h 4. = (3x^2h +3xh^2 +6xh +3h +3h^2 +h^3)/h = 3x^2 +3xh +6x +3 +3h + h^2
Question:function is
f (x) = 1/x.
difference quotient is
[ f (x)  f (a) ] / (x  a)
Now I understand you merely use the first function and plug it in changing the x to an a with f(a). So I did that but I think when simplifying my algebra might have been a little off. If you could help I'd truly appreciate it.
Answers:We know that: f(x) = 1/x f(a) = 1/a Then: [f(x)  f(a)] / (x  a) = (1/x  1/a) / (x  a) = (a  x) / [ax(x  a)] (multiply by xa/xa) = (x  a) / [ax(x  a)] = 1/ax <== ANSWER I hope this helps!
Answers:We know that: f(x) = 1/x f(a) = 1/a Then: [f(x)  f(a)] / (x  a) = (1/x  1/a) / (x  a) = (a  x) / [ax(x  a)] (multiply by xa/xa) = (x  a) / [ax(x  a)] = 1/ax <== ANSWER I hope this helps!
Question:Find the difference quotient for the function f(x)=9x+8, and simplify it.
I don't know. That's why I'm here and trying to figure this one out. I don't get it. Well, I'm at a community college right now. Maybe that's part of the problem??
This is an entirely new concept to me...I am an arts major and never have been mathematically/scientifically inclined  if there was a way both math & science could be better incorporated into 'everyday' situations (in my opinion they are not), then I think I could understand this stuff a little better! But that's just me. :)
Answers:Use (f(x+h)  f(x))/h so... (9 (x + h) + 8  (9x + 8)) / h = (9x + 9h + 8  9x  8) / h = 9h / h = 9
Answers:Use (f(x+h)  f(x))/h so... (9 (x + h) + 8  (9x + 8)) / h = (9x + 9h + 8  9x  8) / h = 9h / h = 9
From Youtube
The Difference Quotient and Extrema Part 1 :I show you how the difference quotient was derived and also begin to explain the concept of extrema. Check out PatrickJMT and MathMikie.
Product and Quotient Laws of Logarithms Part1 :This lesson shows how to use the Product and Quotient Laws of Logarithms to simplify (and also how to) expand logarithmic expressions. This is the first part of a two part lesson. This lesson was created for the MHF4U Advanced Functions course in the province of Ontario, Canada.