1. ## Help!! Difference Quotients.

Find the difference quotient of f, that is, find [f(x + h) - f(x)]/h, where h cannot = 0, for the given function.

Be sure to simplify.

f(x) = (x+3)^2
f(x)=3x^2-5x+4
g(x)=x-3

NOTE: This question comes from a pre-calculus textbook and so the explanation must be in pre-calculus terms and NOT calculus.

Thankx.

2. Originally Posted by AbbbyyJackson
Find the difference quotient of f, that is, find [f(x + h) - f(x)]/h, where h cannot = 0, for the given function.

Be sure to simplify.

f(x) = (x+3)^2
f(x)=3x^2-5x+4
g(x)=x-3

NOTE: This question comes from a pre-calculus textbook and so the explanation must be in pre-calculus terms and NOT calculus.

Thankx.

$\displaystyle \frac{(x+h+3)^2-(x+3)^2}{h}$

$\displaystyle =\frac{[(x+3)+h]^2-(x=3)^2}{h}$

$\displaystyle =\frac{(x+3)^2+2h(x+3)+h^2-(x+3)^2}{h}$

$\displaystyle =\frac{2h(x+3)+h^2}{h}$

$\displaystyle =2(x+3)+h$

$\displaystyle \frac{(x+h+3)^2-(x+3)^2}{h}$

$\displaystyle =\frac{[(x+3)+h]^2-(x=3)^2}{h}$

$\displaystyle =\frac{(x+3)^2+2h(x+3)+h^2-(x+3)^2}{h}$

$\displaystyle =\frac{2h(x+3)+h^2}{h}$

$\displaystyle =2(x+3)+h$
Thankx.
But Can You Explain What You Did.

4. Originally Posted by AbbbyyJackson
Thankx.
But Can You Explain What You Did.
Sure, If we have $\displaystyle f(x)=(x+3)^2$ then we simply replace the $\displaystyle x$ with whatever value is in the parenthesis on the right side. For example $\displaystyle f(2)=(2+3)^2$, or $\displaystyle f(a)=(a+3)^2$. So what if I want to input $\displaystyle x+h$? What I do is replace the $\displaystyle x$ in $\displaystyle (x+3)^2$ with $\displaystyle x+h$. Therefore:

$\displaystyle f(x+h)=[(x+h)+3)]^2$

The numerator in the difference quotient is $\displaystyle f(x+h)-f(x)$. We substitute what we know:

$\displaystyle f(x+h)-f(x)=[(x+h)+3]^2-(x+3)^2$.

The $\displaystyle (x+3)^2$ suggest that we should rearrange the first term in the numerator.

$\displaystyle [(x+3)+h]^2-(x+3)^2$. Notice that the first term is just a binomial. Therefore we can take it's square as $\displaystyle (x+3)^2+2h(x+3)+h^2$

The numerator becomes:

$\displaystyle (x+3)^2+2h(x+3)+h^2-(x+3)^2$

Notice the cancellation of the $\displaystyle (x+3)^2$ term:

$\displaystyle =2h(x+3)+h^2$

The difference quotient is now:

$\displaystyle \frac{2h(x+3)+h^2}{h}$

The $\displaystyle h$ in the denominator cancels:

$\displaystyle =2(x+3)+h$