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If f(x) = 3x + 3/x, then [f(x + h) - f(x)]/h =

 Apr 24, 2014

Best Answer 

 #5
avatar+890 
+5

I've been outside cleaning my car Melody, a seagull had made a mess (and you wouldn't believe the extent of the mess) of the roof and windscreen.

As to LaTex I'm certainly not an expert, I just tend to muddle my way through. Muddling my way through this one, the best I can come up with is

\displaystyle \begin{array}{lr} \text{ lim }\\ h \rightarrow 0 \\  \end{array} \frac{3x+h}{x+h}.

That produces

$$\displaystyle \begin{array}{lr} \text{ lim }\\ h \rightarrow 0 \\ \end{array} \frac{3x+h}{x+h}$$

I've tried to make the h arrow 0 smaller, but without success.

 Apr 25, 2014
 #1
avatar+111438 
+5

If f(x) = 3x + 3/x, then [f(x + h) - f(x)]/h =

----------------------------------------------------------------------------------------------------------------------------

I'm making a broad assumption here, but in my experience,  I believe you're trying to ultimately find the "difference quotient" ⇒  (the derivative of f(x) )

Let's simplify f(x)   so we have

f(x) = [(3x2 + 3)/ x ] ......  then [f(x + h) - f(x)]/h =

([(3(x+h)2 + 3) / (x+h)] - [(3x2 + 3)/ x ]) / h   =

([3x2 + 6xh + 3h2 + 3] / (x + h)] - [(3x2 + 3)/ x ]) / h  =

[3x3 + 6x2h + 3xh2 + 3x - 3x3 - 3x2h -3x - 3h) / [(x)(x+h)(h)] =

 [ 3x2h + 3xh2 - 3h) / [(x)(x+h)(h)] =

(3x2 ) / (x2 + xh)    + 3h / (x+h) - 3 / [(x+h) (x)]

Now, let's take the limit of everything by letting h ⇒ 0      We get

3 - 3x-2

Note that this is just the derivative of f(x)      (Which is what we were hoping for!!!)  

 Apr 24, 2014
 #2
avatar+110163 
+5

OR maybe this was the question

$$f(x)=3x+\frac{3}{x}=\frac{3x^2+3}{x}\\\\
f(x+h)=3(x+h)+\frac{3}{x+h}\\\\
f(x+h)-f(x)=3x+3h+\frac{3}{x+h}-3x-\frac{3}{x}\\\\
f(x+h)-f(x)=3h+\frac{3}{x+h}-\frac{3}{x}\\\\
f(x+h)-f(x)=\frac{3h(x+h)x+3x-3(x+h)}{x(x+h)}\\\\
f(x+h)-f(x)=\frac{3h(x+h)x-3h}{x(x+h)}\\\\
\left[f(x+h)-f(x)\right]\div h=\frac{3(x+h)x-3}{x(x+h)}\\\\$$

Now taking it the extra logical step

$$\left[f(x+h)-f(x)\right]\div h=\frac{3(x+h)x-3}{x(x+h)}\\\\$$

limit as h tends to 0 =   $$\frac{3x^2-3}{x^2}$$

Now, if I didn't make any mistakes this is the derivative of the original equation. And it is!   

 

NOW can someone tell me how to write limit as h tends to 0 in latex please?

 Apr 25, 2014
 #3
avatar+30637 
+5

 NOW can someone tell me how to write limit as h tends to 0 in latex please?

The following:  

$$\lim h\rightarrow0\frac{3x+h}{x+h}$$ (in Texmaker go to Math/Math functions to get \lim)gives:

$$$$\lim h\rightarrow0\frac{3x+h}{x+h}$$$$

but I don't know how to get the h->0 to go just beneath lim. We need a LaTeX expert!
 Apr 25, 2014
 #4
avatar+110163 
0

Where is Rom and Bertie when you need them?

 Apr 25, 2014
 #5
avatar+890 
+5
Best Answer

I've been outside cleaning my car Melody, a seagull had made a mess (and you wouldn't believe the extent of the mess) of the roof and windscreen.

As to LaTex I'm certainly not an expert, I just tend to muddle my way through. Muddling my way through this one, the best I can come up with is

\displaystyle \begin{array}{lr} \text{ lim }\\ h \rightarrow 0 \\  \end{array} \frac{3x+h}{x+h}.

That produces

$$\displaystyle \begin{array}{lr} \text{ lim }\\ h \rightarrow 0 \\ \end{array} \frac{3x+h}{x+h}$$

I've tried to make the h arrow 0 smaller, but without success.

Bertie Apr 25, 2014
 #6
avatar+110163 
0

\displaystyle \begin{array}{lr} \text{ lim }\\ h \rightarrow 0 \\  \end{array} \frac{3x+h}{x+h}.

That produces

$$\displaystyle \begin{array}{lr} \text{ lim }\\ h \rightarrow 0 \\ \end{array} \frac{3x+h}{x+h}$$

 

Thanks Bertie,

What is the {lr} for?

 

I wrote a  post today that mentioned seagulls, just as well it wasn't one of those giving your car grief!

http://web2.0calc.com/questions/this-problem-was-on-my-homework-last-night-and-i-just-don-t-know-how-i-m-supposed-to-solve-it

 Apr 25, 2014
 #7
avatar+890 
0

The {lr} doesn't do anything, I was thinking of something else and forgot to scrub it .

For an array with two columns, it would align the first column to the left and the second to the right.

 Apr 25, 2014
 #8
avatar+110163 
0

Thank you Bertie

 Apr 25, 2014

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