QuestionFind the difference quotient $\frac{f(x+h)-f(x)}{h}$ for $f(x)=9x+1$ and $h \neq 0$ . Simplify your answer.

Studdy Solution

STEP 1

Assumptions1. The function is $f(x)=9x+1$ . We are asked to find the difference quotient $\frac{f(x+h)-f(x)}{h}$ for $h \neq0$

STEP 2

First, we need to calculate $f(x+h)$ . This is done by replacing $x$ in the function $f(x)$ with $(x+h)$ .
$f(x+h) =9(x+h) +1$

STEP 3

implify the expression for $f(x+h)$ .
$f(x+h) =9x +9h +1$

STEP 4

Now, we can substitute $f(x+h)$ and $f(x)$ into the difference quotient formula.
$\frac{f(x+h)-f(x)}{h} = \frac{(9x +9h +1) - (9x +1)}{h}$

STEP 5

implify the numerator of the difference quotient.
$\frac{f(x+h)-f(x)}{h} = \frac{9h}{h}$

STEP 6

Finally, simplify the difference quotient by cancelling out the $h$ in the numerator and the denominator.
$\frac{f(x+h)-f(x)}{h} =9$ So, the difference quotient for the function $f(x)=9x+1$ is9.

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QuestionFind the difference quotient $\frac{f(x+h)-f(x)}{h}$ for $f(x)=9x+1$ and $h \neq 0$ . Simplify your answer.

Studdy Solution

STEP 1

Assumptions1. The function is $f(x)=9x+1$ . We are asked to find the difference quotient $\frac{f(x+h)-f(x)}{h}$ for $h \neq0$

STEP 2

First, we need to calculate $f(x+h)$ . This is done by replacing $x$ in the function $f(x)$ with $(x+h)$ .
$f(x+h) =9(x+h) +1$

STEP 3

implify the expression for $f(x+h)$ .
$f(x+h) =9x +9h +1$

STEP 4

Now, we can substitute $f(x+h)$ and $f(x)$ into the difference quotient formula.
$\frac{f(x+h)-f(x)}{h} = \frac{(9x +9h +1) - (9x +1)}{h}$

STEP 5

implify the numerator of the difference quotient.
$\frac{f(x+h)-f(x)}{h} = \frac{9h}{h}$

STEP 6

Finally, simplify the difference quotient by cancelling out the $h$ in the numerator and the denominator.
$\frac{f(x+h)-f(x)}{h} =9$ So, the difference quotient for the function $f(x)=9x+1$ is9.

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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)​ for f(x)=9x+1f(x)=9x+1f(x)=9x+1 and h≠0h \neq 0h=0. Simplify your answer.

Studdy Solution

STEP 1

Assumptions1. The function is f(x)=9x+1f(x)=9x+1f(x)=9x+1 . We are asked to find the difference quotient f(x+h)−f(x)h\frac{f(x+h)-f(x)}{h}hf(x+h)−f(x)​ for h≠0h \neq0h=0

STEP 2

First, we need to calculate f(x+h)f(x+h)f(x+h). This is done by replacing xxx in the function f(x)f(x)f(x) with (x+h)(x+h)(x+h).f(x+h)=9(x+h)+1f(x+h) =9(x+h) +1f(x+h)=9(x+h)+1

STEP 3

implify the expression for f(x+h)f(x+h)f(x+h).f(x+h)=9x+9h+1f(x+h) =9x +9h +1f(x+h)=9x+9h+1

STEP 4

Now, we can substitute f(x+h)f(x+h)f(x+h) and f(x)f(x)f(x) into the difference quotient formula.f(x+h)−f(x)h=(9x+9h+1)−(9x+1)h\frac{f(x+h)-f(x)}{h} = \frac{(9x +9h +1) - (9x +1)}{h}hf(x+h)−f(x)​=h(9x+9h+1)−(9x+1)​

STEP 5

implify the numerator of the difference quotient.f(x+h)−f(x)h=9hh\frac{f(x+h)-f(x)}{h} = \frac{9h}{h}hf(x+h)−f(x)​=h9h​

STEP 6

Finally, simplify the difference quotient by cancelling out the hhh in the numerator and the denominator.f(x+h)−f(x)h=9\frac{f(x+h)-f(x)}{h} =9hf(x+h)−f(x)​=9So, the difference quotient for the function f(x)=9x+1f(x)=9x+1f(x)=9x+1 is9.

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QuestionFind the difference quotient $\frac{f(x+h)-f(x)}{h}$ for $f(x)=9x+1$ and $h \neq 0$ . Simplify your answer.

Assumptions1. The function is $f(x)=9x+1$ . We are asked to find the difference quotient $\frac{f(x+h)-f(x)}{h}$ for $h \neq0$

First, we need to calculate $f(x+h)$ . This is done by replacing $x$ in the function $f(x)$ with $(x+h)$ .
$f(x+h) =9(x+h) +1$

implify the expression for $f(x+h)$ .
$f(x+h) =9x +9h +1$

Now, we can substitute $f(x+h)$ and $f(x)$ into the difference quotient formula.
$\frac{f(x+h)-f(x)}{h} = \frac{(9x +9h +1) - (9x +1)}{h}$

implify the numerator of the difference quotient.
$\frac{f(x+h)-f(x)}{h} = \frac{9h}{h}$

Finally, simplify the difference quotient by cancelling out the $h$ in the numerator and the denominator.
$\frac{f(x+h)-f(x)}{h} =9$ So, the difference quotient for the function $f(x)=9x+1$ is9.