Differntiable at point problem

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f(x) = x/(3x + 1), prove f(x) is differentiable at point 2.

Ok so I've had several attempts at this...

Using Q(h) = (f(h) - f(2))/h

I eventually end up with (h^2 -2h)/(7(3h + 1))

Obviously the above is rubbish because it I differentiate f(x) using the normal rules then

dy/dx = 1/(3x + 1)^2

What am I missing here?

Also I've tried using the difference quotient to prove it is differentiable but same result - just rubbish.

f(c + h) - f(c)/ h

The above also doesn't work out either! Please hep!
 
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So you have the function [itex]f(x)=\frac{x}{3x+1}[/itex] and you know that a function is differentiable at [itex]a[/itex] if its derivative exists at [itex]a[/itex]. You also know that

[tex]\left.\frac{df}{dx}\right|_{a}\equiv \lim_{h\rightarrow 0}\frac{f(a+h)-f(a)}{h}=\frac{\frac{a+h}{3(a+h)+1}-\frac{a}{3a+1}}{h}[/tex]

If you simplify this does it match what you expected by using the rules you know? When you take the limit does that prove the limit exists for [itex]a=2[/itex]?
 
james.farrow said:
f(x) = x/(3x + 1), prove f(x) is differentiable at point 2.

Ok so I've had several attempts at this...

Using Q(h) = (f(h) - f(2))/h
It would be better to use Q(h)= (f(2+ h)- f(2))/h!

I eventually end up with (h^2 -2h)/(7(3h + 1))
Obviously the above is rubbish
Yes, because you used the wrong formula for the difference quotient.

because it I differentiate f(x) using the normal rules then

dy/dx = 1/(3x + 1)^2

What am I missing here?

Also I've tried using the difference quotient to prove it is differentiable but same result - just rubbish.

f(c + h) - f(c)/ h

The above also doesn't work out either! Please hep!
 
Jefferydk if I simplify it I end up with nothing like? Whats going on...?
 
Err after a cup of tea and a break the penny drops...

Thanks for your help lads! No doubt I'll call on you again!

James
 
A cup of tea works wonders!

(I used to do my best work after a couple glasses of rum- but then I could never find the papers where I had written it all down!)
 

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