Math / Algebra

QuestionFind the difference quotient $\frac{f(x+h)-f(x)}{h}$ , where $h \neq 0$ , $f(x)=8 x-2$
Simplify your answer as much as possible. $\frac{f(x+h)-f(x)}{h}=$

Studdy Solution

STEP 1

What is this asking? We're finding how much the function $f(x)$ changes on average over a small interval of size $h$ . Watch out! Don't forget to distribute the negative sign correctly when subtracting $f(x)$ !

STEP 2

1. Substitute the function
2. Expand and simplify

STEP 3

Alright, let's kick things off by substituting the expression for $f(x+h)$ into our difference quotient formula!
Remember, $f(x) = 8x - 2$ , so $f(x+h) = 8(x+h) - 2$ .
So we get: $\frac{f(x+h) - f(x)}{h} = \frac{8(x+h) - 2 - f(x)}{h}$

STEP 4

Now, let's plug in our expression for $f(x)$ , which is $8x - 2$ : $\frac{8(x+h) - 2 - (8x - 2)}{h}$ Notice those parentheses around $8x - 2$ !
They're super important because we're subtracting the entire function $f(x)$ .

STEP 5

Let's distribute that 8 in the numerator: $\frac{8x + 8h - 2 - (8x - 2)}{h}$ Now, distribute the negative sign: $\frac{8x + 8h - 2 - 8x + 2}{h}$ Look closely!
We have $8x$ and $-8x$ , and $-2$ and $+2$ .
These will add to zero, which simplifies things nicely: $\frac{8h}{h}$

STEP 6

Since $h \neq 0$ , we can divide the numerator and denominator by $h$ , which is the same as multiplying by $\frac{1}{h}$ in the numerator and $\frac{1}{h}$ in the denominator: $\frac{8h}{h} = \frac{8h}{1 \cdot h} = \frac{8h \cdot \frac{1}{h}}{1 \cdot h \cdot \frac{1}{h}} = \frac{8 \cdot 1}{1 \cdot 1} = 8$ So, our final simplified expression is $8$ !

STEP 7

$\frac{f(x+h)-f(x)}{h}=8$

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QuestionFind the difference quotient $\frac{f(x+h)-f(x)}{h}$ , where $h \neq 0$ , $f(x)=8 x-2$
Simplify your answer as much as possible. $\frac{f(x+h)-f(x)}{h}=$

Studdy Solution

STEP 1

What is this asking? We're finding how much the function $f(x)$ changes on average over a small interval of size $h$ . Watch out! Don't forget to distribute the negative sign correctly when subtracting $f(x)$ !

STEP 2

1. Substitute the function
2. Expand and simplify

STEP 3

Alright, let's kick things off by substituting the expression for $f(x+h)$ into our difference quotient formula!
Remember, $f(x) = 8x - 2$ , so $f(x+h) = 8(x+h) - 2$ .
So we get: $\frac{f(x+h) - f(x)}{h} = \frac{8(x+h) - 2 - f(x)}{h}$

STEP 4

Now, let's plug in our expression for $f(x)$ , which is $8x - 2$ : $\frac{8(x+h) - 2 - (8x - 2)}{h}$ Notice those parentheses around $8x - 2$ !
They're super important because we're subtracting the entire function $f(x)$ .

STEP 5

Let's distribute that 8 in the numerator: $\frac{8x + 8h - 2 - (8x - 2)}{h}$ Now, distribute the negative sign: $\frac{8x + 8h - 2 - 8x + 2}{h}$ Look closely!
We have $8x$ and $-8x$ , and $-2$ and $+2$ .
These will add to zero, which simplifies things nicely: $\frac{8h}{h}$

STEP 6

STEP 7

$\frac{f(x+h)-f(x)}{h}=8$

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QuestionFind the difference quotient f(x+h)−f(x)h\frac{f(x+h)-f(x)}{h}hf(x+h)−f(x)​, where h≠0h \neq 0h=0, f(x)=8x−2f(x)=8 x-2f(x)=8x−2Simplify your answer as much as possible. f(x+h)−f(x)h=\frac{f(x+h)-f(x)}{h}=hf(x+h)−f(x)​=

Studdy Solution

STEP 1

What is this asking? We're finding how much the function f(x)f(x)f(x) changes on average over a small interval of size hhh. Watch out! Don't forget to distribute the negative sign correctly when subtracting f(x)f(x)f(x)!

STEP 2

1. Substitute the function2. Expand and simplify

STEP 3

STEP 4

STEP 5

STEP 6

STEP 7

f(x+h)−f(x)h=8\frac{f(x+h)-f(x)}{h}=8hf(x+h)−f(x)​=8

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Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionFind the difference quotient $\frac{f(x+h)-f(x)}{h}$ , where $h \neq 0$ , $f(x)=8 x-2$
Simplify your answer as much as possible. $\frac{f(x+h)-f(x)}{h}=$

What is this asking? We're finding how much the function $f(x)$ changes on average over a small interval of size $h$ . Watch out! Don't forget to distribute the negative sign correctly when subtracting $f(x)$ !

1. Substitute the function
2. Expand and simplify

$\frac{f(x+h)-f(x)}{h}=8$