solving the expression

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May 12th 2008, 12:13 AM
lemontea

solving the expression

Let function NoFontMessage () {} jsMath = {Parser: {prototype: {environments: {'eqnarray*' :['Array',null,null,'rcl',[5/18,5/18],3,'D']}}}}; g(x)=1/(x+2) Find the expression

g(x+h)-g(x) / h

this is what i did:
g(x+h)-g(x) / h
=g(x)+g(h)-g(x)/ h
=1/(x+2)+g(h)-1/(x+2) / h
=g(h) / h
= g

is this right?
May 12th 2008, 12:26 AM
earboth

Quote:

Originally Posted by lemontea

Let g(x)=1/(x+2)
Find the expression

g(x+h)-g(x) / h

this is what i did:
g(x+h)-g(x) / h
=g(x)+g(h)-g(x)/ h
=1/(x+2)+g(h)-1/(x+2) / h
=g(h) / h
= g

is this right?

Unfortunately no.

$g(x)=\frac1{x+2}~\implies~g(x+h)=\frac1{x+2+h}$

$ \frac{\frac1{x+2+h} - \frac1{x+2}}{h} = \frac{\frac{x+2-(x+2+h)}{(x+2)(x+2+h)}}{h} = \frac{\frac{-h}{(x+2)(x+2+h)}}{h} = -\frac1{(x+2)(x+2+h)}$
May 12th 2008, 12:28 AM
Moo

Hello,

Quote:

Originally Posted by lemontea

Let function NoFontMessage () {} jsMath = {Parser: {prototype: {environments: {'eqnarray*' :['Array',null,null,'rcl',[5/18,5/18],3,'D']}}}}; g(x)=1/(x+2) Find the expression

g(x+h)-g(x) / h

this is what i did:
g(x+h)-g(x) / h
=g(x)+g(h)-g(x)/ h
=1/(x+2)+g(h)-1/(x+2) / h
=g(h) / h
= g

is this right?

Erm.. I don't know what to say... In which grade are you ?

$g(x+h) {\color{red}\neq} g(x)+g(h)$

And $\frac{g(h)}{h} {\color{red}\neq} g$ (following your reasoning, it'd make g() lol)...

You want $\frac{g(x+h)-g(x)}{h}$

$g(x)=\frac{1}{x+2}$

--> $g({\color{blue}x+h})=\frac{1}{({\color{blue}x+h})+ 2}=\frac{1}{x+h+2}$

$\frac{g(x+h)-g(x)}{h} = \frac{\frac{1}{x+h+2}-\frac{1}{x+2}}{h}$

$\begin{aligned} \frac{g(x+h)-g(x)}{h} &= \frac{\frac{1}{x+h+2} \cdot \frac{x+2}{x+2}-\frac{1}{x+2} \cdot \frac{x+h+2}{x+h+2}}{h} \\ \\ &= \frac{\frac{x+2}{(x+h+2)(x+2)}-\frac{x+h+2}{(x+2)(x+h+2)}}{h} \\ &= \frac{\frac{(x+2)-(x+h+2)}{(x+h+2)(x+2)}}{h} \end{aligned}$

$\begin{aligned} \frac{g(x+h)-g(x)}{h} &= \frac{\frac{-h}{(x+h+2)(x+2)}}{h} \\ \\ &= \frac{-{\color{red}h}}{{\color{red}h}(x+h+2)(x+2)} \\ &= \frac{-1}{(x+h+2)(x+2)} \end{aligned}$