Guys, I'm feeling like a brainlet right now. Help would

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Did you lose your cheat sheet?
Here, you can use mine

\[f(x) = \frac{1}{x-3}

\frac{f(x)-f(x_0}{x-x_0} = \frac{\frac{1}{x-3} - \frac{1}{x_0 - 3}}{x - x_0} =

= \frac{\frac{x_0 - 3 - x + 3}{(x-3)(x_0-3)}}{x-x_0} =

= -\frac{x - x_0}{(x-x_0)(x-3)(x_0-3)}

For x \neq x_0

= -\frac{1}{(x-3)(x_0-3)} (1)

Now, as x \rightarrow x_0

(1) \rightarrow -\frac{1}{(x_0-3)^2}

So, f'(x_0) = -\frac{1}{(x_0-3)^2} \]

>>9150261
I'm having some trouble trying to read that.

>>9150260
Cool cheat sheet. Is this for philosophy or something?

>>9150260
Are buddhists supposed to memorize this?

>>9150268
Yeah, it's for my Applied Logic 101 class

>>9150266
\[f(x) = \frac{1}{x-3} \]

\[ \frac{f(x)-f(x_0}{x-x_0} = \frac{\frac{1}{x-3} - \frac{1}{x_0 - 3}}{x - x_0} = \frac{\frac{x_0 - 3 - x + 3}{(x-3)(x_0-3)}}{x-x_0} = -\frac{x - x_0}{(x-x_0)(x-3)(x_0-3)} (1) \]

For $x \neq x_0$:

\[ (1) = -\frac{1}{(x-3)(x_0-3)} (2) \]

Now, as $x \rightarrow x_0$:

\[ (2) \rightarrow -\frac{1}{(x_0-3)^2} \]

So, $f'(x_0) = -\frac{1}{(x_0-3)^2}$

>>9150244
Is this a calc1 question or something like that? I'm reading this as what is the derivative of f(x) at 3 and (the other problem) the derivative of f(x) at -2. Is this correct or are you asking a question which involves something more of the nature of logic or number theory?

>>9150244
>>9150244
Holy shit I've done that at the start of the uni and I don't remember shit about that right now.

>>9150244
you have to compute the limit.. look at the definition for a derivative

also
R/{3} means R minus the point 3 (which is a singularity for f, IE f(x) is not defined for x=3)
likewise for the other one

>>9150407
>>9150261
my browser isn't formatting this into mathematese

>>9150431
It's them not your browser

lim f(x+h) - f(x)
h--> 0

>>9150407
Use [ eqn] and [ math] tags instead, newfriend:

[eqn]f(x) = \frac{1}{x-3} [/eqn][eqn] \frac{f(x)-f(x_0}{x-x_0} = \frac{\frac{1}{x-3} - \frac{1}{x_0 - 3}}{x - x_0} = \frac{\frac{x_0 - 3 - x + 3}{(x-3)(x_0-3)}}{x-x_0} = -\frac{x - x_0}{(x-x_0)(x-3)(x_0-3)} (1) [/eqn]For [math]x \neq x_0[/math]:[eqn] (1) = -\frac{1}{(x-3)(x_0-3)} (2) [/eqn] Now, as [math]x \rightarrow x_0[/math]: [eqn] (2) \rightarrow -\frac{1}{(x_0-3)^2} [/eqn]So, [math]f'(x_0) = -\frac{1}{(x_0-3)^2}[/math]