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Compute the difference quotient $\frac{f(x+h)-f(x)}{h}, \quad h \neq 0$ Whenever possible, simplify the expression so that the resulting expression is defined when $h=0$.$f(x)=2 x^{2}+3 x-7$

$$4 x+2 h+3$$

Algebra

Chapter 1

Functions and their Applications

Section 2

Basic Notions of Functions

Functions

Harvey Mudd College

University of Michigan - Ann Arbor

Idaho State University

Lectures

01:43

In mathematics, a function is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output. An example is the function that relates each real number x to its square x^2. The output of a function f corresponding to an input x is denoted by f(x).

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Compute the difference quo…

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Find and simplify the diff…

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for this problem. We've been given a function f of X equals two x squared plus three X minus seven, and we want to compute the difference quotient f of X plus h minus F of X over H, and this quotient is one that we'll see often throughout calculus. So when we go to evaluate this, you know it's our numerator has two different um calling is to that function F. The second one is just plain old F of X on the first one is F of X plus h, which means I'll be substituting X plus h in for where X is in my original function. So let's, um, let's put those numbers in. First of all, I have to times. Well, here's my first substitution. X plus H squared plus three Substitute again X plus A H minus seven, and I want to subtract just plain old F of X, which is two X squared plus three x minus seven. We are subtracting that entire piece that don't forget to put it in parentheses. We have to distribute that subtraction sign and it's all over H. Now let's get rid of our parentheses. Let's square X plus h, and we're gonna multiply everything by two. So that would give me two X squared. Normally, it would be two x h times, too. That's four x h plus two h squared plus three X plus three H minus seven. Distribute that subtraction minus two X squared minus three X plus seven all over H. Now there's a lot here, but fortunately we have a lot that cancels. I have two X squared minus two x squared three X minus three X and negative seven plus seven. So, after I've gotten rid of all of that, what I have left is four x h plus two h squared plus three h over H. Although every single term in the numerator has an h so I can cancel those h is with the H and the denominator, and that leaves me with four x plus two h plus three.

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