# Limit of Average Rate of Change

• Jan 18th 2009, 08:29 AM
tom ato
Limit of Average Rate of Change
How do I find the instantaneous rate of change when

f(x)=3x-4; x=2

I know that the limit as h->0 is

[f(x+h) - f(x)]/h

but I'm not sure how to plug in the 3x-4. I was able to find the limit when f(x) = x^2 just by [(x+h)^2 - x^2]/h, but I can't seem to find the correct way to get 3x-4 into the equation (I keep ending up w/a 0 in the denominator no matter what I try).
• Jan 18th 2009, 08:33 AM
Mush
Quote:

Originally Posted by tom ato
How do I find the instantaneous rate of change when

f(x)=3x-4; x=2

I know that the limit as h->0 is

[f(x+h) - f(x)]/h

but I'm not sure how to plug in the 3x-4. I was able to find the limit when f(x) = x^2 just by [(x+h)^2 - x^2]/h, but I can't seem to find the correct way to get 3x-4 into the equation (I keep ending up w/a 0 in the denominator no matter what I try).

It doesn't really matter what x is.

The first differential of a linear function like yours is a constant, which is independent of x. The instantaneous time rate of change is the same as all other time rates of change. The answer is 3.

If you're asking how to find the rate of change by first principles then it's this:

If $\displaystyle f(x) = 3x-4$

Then $\displaystyle f(x+h)$ is the exact same equation as above, only instead of x, you have x+h, so:

$\displaystyle f(x+h) = 3(x+h)-4 = 3x+3h-4$

Applying this to the definition of a first differential:

$\displaystyle f'(x) = \displaystyle \lim_{h \to 0} \frac{3(x+h)-4-(3x-4)}{h}$

$\displaystyle = \displaystyle \lim_{h \to 0} \frac{3x+3h-4-3x+4}{h}$

$\displaystyle = \displaystyle \lim_{h \to 0} \frac{3h}{h}$

$\displaystyle = \displaystyle \lim_{h \to 0} 3$

$\displaystyle = \displaystyle 3$

Or by plugging in x = 2:

$\displaystyle f'(x) = \displaystyle \lim_{h \to 0} \frac{3(2+h)-4-(3(2)-4)}{h}$

$\displaystyle = \displaystyle \lim_{h \to 0} \frac{6+3h-4-6+4}{h}$

$\displaystyle = \displaystyle \lim_{h \to 0} \frac{3h}{h}$

$\displaystyle = \displaystyle \lim_{h \to 0} 3$

$\displaystyle = \displaystyle 3$
• Jan 18th 2009, 08:34 AM
running-gag
Hi

I don't see where you get trouble

$\displaystyle \frac{f(x+h) - f(x)}{h} = \frac{[3(x+h)-4]-[3x-4]}{h} = \frac{3h}{h} = 3$

EDIT : sorry Mush
• Jan 18th 2009, 08:54 AM
tom ato
Thanks to both. I forgot to distribute the negative at -(3x-4)...
• Jan 18th 2009, 09:01 AM
Mush
Quote:

Originally Posted by tom ato
Thanks to both. I forgot to distribute the negative at -(3x-4)...

With elementary functions like these, it's easier to differentiate them first, the way you normally would. That way, once you apply your differentiation by first principles, you know what your goal is, and that can help you identify any mistakes.