how to solve this math problem involving precalculus?

Question

given f(s)=4s^2+3s+8
find [f(t+h)-f(t)]\h

Answer 1

This is more of a college algebra problem than a calculus problem, but here goes.

f(t+h) = 4(t+h)^2+3(t+h)+8
f(t+h) = 4t^2 + 8th + 4h^2 + 3t + 3h +8

f(t) = 4t^2+3t+8

So f(t+h) - f(h) = (4t^2 + 8th + 4h^2 + 3t + 3h +8) - (4t^2+3t+8)
f(t+h) - f(h) = 4t^2 + 8th + 4h^2 + 3t + 3h +8 - 4t^2 - 3t - 8
f(t+h) - f(h) = 8th + 4h^2 + 3h

Then [f(t+h) - f(h)]/h = (8th + 4h^2 + 3h)/h
[f(t+h) - f(h)]/h = 8t + 4h + 3

Hope this helps!

Answer 2

To find [ f(t + h) - f(t) ] / h, change the variable from t to s.

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

= [ 4(s + h)^2 + 3(s + h) + 8 - ( 4s^2 + 3s + 8)] / h

= [ 4(s^2 + 2sh + h^2) + 3s + 3h + 8 - 4s^2 - 3s - 8 ] / h

= [ 4s^2 + 8sh + 4h^2 + 3s + 3h + 8 - 4s^2 - 3s - 8 ] / h

= [ 8sh + 4h^2 + 3h ] / h

= 8s + 4h + 3

If the variable t is desired, then change the variable from t to s.

8t + 4h + 3

Answer 3

I'll tell you how to solve, but not the solution (that would be cheating).

Since the problem doesn't tell you how much "t" and "h" are, the best you can do is give the answer as an expression with t's and h's in it; you can't give the answer as a number.

To write an expression for f(t+h), use the definition of f(s), and plug in "t+h" wherever you see an "s".

To write an expression for f(h), use the definition of f(s), and plug in "h" wherever you see an "s".

Then, you need to subtract one of those expressions from the other and divide by h. (That's because you're asked to find [f(t+h)-f(t)]\h )

At that point you'll have one big expression with t's and h's in it. You'll probably be able to simplify it a little (combine like terms, etc.) to make it look a little less hairy.

Answer 4

properly, your question must be seen a rant, that's against community rules. So, if somebody considerably disagreed with you, or in the event that they have been basically a stickler for the guidelines, they might justifiably record you and characteristic your question bumped off. As on your question, i'm able to basically speculate. you may desire to bear in mind that at school, arithmetic is taught basically on the least education for existence, so rote-found out suggestions are quite all they budgeted for. Calculus has been suitable for a protracted time to a great many disciplines, while discrete arithmetic has basically been suitable for a quite short quantity of time, and nevertheless to a small style of fields (computing gadget technological know-how, are you able to think of of everywhere else?). quite, the main suitable area of this is enumeration, that's lined in an admittedly cursory way.

Answer 5

put the values you will receive this equation
lim h->0 ( 4(t+h)^2+3(t+h0+8-4t^2-3t-8)/h
open it
4t^2+4h^2+8th+3t+3h+8-4t^2-3t-8/h
(4h^2+8th+3h)/h
cancel h both in nuerator and denominator
u will get
4h+8t+3
put h as 0
answer 8t+3

Answer 6

with this one you simply need to put (t+h) and (t) in for any s in the given equation...like this (forgetting about the denominator for now...

f(s)=(4(t+h)^2+3(t+h)+8)-(4(t)^2+3t+8)

then simplify

(4(t^2+2th+h^2)+3t+3h+8)-(4t^2+3t+8)

and again

4t^2+8th+4h^2+3t+3h+8-4t^2-3t-8

and cancel

8th+4h^2+3h

and simplify

h(8t+4h+3)/(h) (h cancels)

(8t+4h+3)