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Evaluate the difference quotient for the given function. Simplify your answer.

$ f(x) = \dfrac{x + 3}{x + 1} $ , $ \dfrac{f(x) - f(1)}{x - 1} $

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$-\frac{1}{x+1}$

01:57

Jeffrey Payo

02:47

Heather Zimmers

Calculus 1 / AB

Calculus 2 / BC

Calculus 3

Chapter 1

Functions and Models

Section 1

Four Ways to Represent a Function

Functions

Integration Techniques

Partial Derivatives

Functions of Several Variables

Missouri State University

University of Michigan - Ann Arbor

Idaho State University

Boston College

Lectures

04:31

A multivariate function is a function whose value depends on several variables. In contrast, a univariate function is a function whose value depends on only one variable. A multivariate function is also called a multivariate expression, a multivariate polynomial, a multivariate series, or a multivariate function of several variables.

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In calculus, partial derivatives are derivatives of a function with respect to one or more of its arguments, where the other arguments are treated as constants. Partial derivatives contrast with total derivatives, which are derivatives of the total function with respect to all of its arguments.

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Evaluate the difference qu…

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Find and simplify the diff…

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02:36

All right here we have a function in blue, it's f of x, equals x, plus 3 over x, plus 1, and our goal is defined. The difference quotient in red f of x, minus f of 1 over x minus 1 point so before we do that. Let'S go ahead and find the value of f of 1 time so f of 1 is going to be the same as f of x. Only x is replaced by 1, so we'll get 1 plus 3 in the numerator over 1 plus 1 in the denominator. So that gives us 4 over 2, which reduces down to 2 point okay, so f of 1 is 2, so let's go ahead and keep working with our difference. Potion f of x is given so we'll just rewrite the fraction and then we'll subtract of f of 1, which is simply 2, and we still divide by x, minus 1 point all right. The easiest way to clean up a fraction within a fraction is to multiply top and bottom by the denominator of their fractions, so we're going to go ahead and multiply both top and bottom. Both our numerator and our denominator by x plus 1 are excellent, all right. So when we do that, then we are going to basically distribute the x plus 1 to the fraction, as well as to the minus 2 point. So let's go ahead and distribute it first to the fraction that gives us x, plus 3 over x, plus 1 times x, plus 1, and that's our first term distributed. Then our second term is twice x, plus 1, okay and then the bottom i'm going to leave it in factored form. You'Ll see why and then a little bit. Why don't go ahead and foil? Okay, so let's notice that top and bottom we get something cool in this. This term here notice that we have an x plus 1 term in the numerator and the x plus 1 term in the denominator, so those can be canceled. So, let's see what we have left now we have just that first term is just x plus 3, the second term i'm going to go ahead and distribute the minus 2 to both parts of the x plus 1. That will give me minus 2 x, minus 2 and that's all over x, minus 1 times x, plus 1 point okay, let's kind of clean up the top. We have notice, we have an x and a minus 2 x, so i can subtract that to get minus x we have plus 3 minus 2 point. So that gives me plus 1 and notice- that's kind of similar to what we see in the bottom, but not quite right. So what i'm going to do is factor out a minus sign in from the top and notice, when i do that, i get minus x. Minus 1, in the parentheses that allows us to cancel with that x, minus 1 in the denominator. So this problem really cleans up nicely, so i'm going to go ahead then and cancel the x minus 1 top and bottom, and so therefore we get our final answer. Minus 1 over x plus 1 is our simplified answer for a difference potion all right. Hopefully that helped have an amazing day.

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