What is the Instantaneous Rate of Change at X=2 for the Function F(x) = 4/(x-1)?

wearingthin
Messages
5
Reaction score
0

Homework Statement



Find the irc at X=2

F(x) = \frac{4}{x-1}



Homework Equations



\stackrel{lim}{h\rightarrow0}\frac{F(a+h)-F(a)}{h}



The Attempt at a Solution



I end up with \frac{4}{h+h^{2}} or some other form of an h on the bottom. Is there something else i can do to it??
 
Physics news on Phys.org
\frac{4}{(2+h)-1} - \frac{4}{2-1}

all over h

is how i set it up originally
 
and then

\frac{4(1)}{(1+h)1} - \frac{4(1+4)}{1(1+h)}

still all over h
 
which becomes


\frac{h}{1+h}

all over h

right??
 
You made the mistake in your third post; that 4 in the very top right should be an h

wearingthin said:
and then

\frac{4(1)}{(1+h)1} - \frac{4(1+4)}{1(1+h)}

still all over h

First factor out the four so it's easier to work with\frac{\frac{4}{h+1} - \frac{4}{1}*\frac{h + 1}{h + 1}}{h} = \frac{4\left(\frac{1}{h + 1} - \frac{h + 1}{h + 1}\right)}{h}
then see if you can cancel that h in the denominator.
 
IRC = -4

You're a genius! Thank you


-Matt
 
Thread 'Use greedy vertex coloring algorithm to prove the upper bound of χ'
Hi! I am struggling with the exercise I mentioned under "Homework statement". The exercise is about a specific "greedy vertex coloring algorithm". One definition (which matches what my book uses) can be found here: https://people.cs.uchicago.edu/~laci/HANDOUTS/greedycoloring.pdf Here is also a screenshot of the relevant parts of the linked PDF, i.e. the def. of the algorithm: Sadly I don't have much to show as far as a solution attempt goes, as I am stuck on how to proceed. I thought...
Back
Top