anti-derivative help please!

Question

the limit below represents the derivative of some function in the number a:

lim h-->0 = 4(3+h)^5 - 972

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

h

I need the original function and the number a please.

Answer 1

substitute h=0 and you get 0/0 an indeterminate form

use L'Hopital's rule and take derivatives, separately of numerator and denominator

to get

20(3+h)4 over 1 = 20(3+h)^4

substitute h=0 again to get 20(3)^4 = 1620 = (5/4)(972)

the limit = 1620

not sure what "a" is without more information, although maybe it's related to 0.8 as .8(1620)=972 or 5/4 as (5/4)1620 = 972

this looks similar to the [f(x+h)-f(x)]/h limit = f'(x) = derivative of f(x). Then f(x+h) =4(3+h)^5 and f(x)=972

= 4(3)^5

maybe f(x) = a4(x)5 where a = (5/4)

f(x) = (5/4)x^5. It's possible. But there must be more to this problem than given.

what's puzzling is normally when you try to take the limit of [f(x+h)-f(x)/h as h goes to zero, the h will cancel out, but with this problem, it doesn't cancel, so it suggests something is amiss in the problem, miscopied or a typo in a textbook. IF it's a problem part of a group of similar problems, there may be some implicit assumption not stated. But I'm wrong. IF I'd just worked out (3+h)^5 the h does cancel.

4(3+h)^5 = 4h^5 + 36h^4 + 252h^2 +984h + 1620h. Divide by h and it leaves

a 4th degree polynomial with constant term 1620. let h=0 and you're left with f'(x) = 1620 = limit as h goes to zero.

f(x) = 4x^5, f(3) = 4(3)^5 = 972

a = 4? a is usually used as a coefficient.

Answer 2

If F(x) is the anti-derivative of f(x) then

f(a) = dF/dx(a) = Limit as h→0 of (F(a+h) - F(a))/h

I think you can work out what F(x) and a is from this, as F(a+h) = 4(3+h)⁵ and 972 = F(a)

You can check to see if you get 972 when you plug in the a value to the function from the 1st term.

Good luck!