anonymous
  • anonymous
For position function s(t):=20-5t^2, if h is a small number, find the average rate of change of s(t) on the interval [2,2+h]. (Simplify fully)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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amistre64
  • amistre64
ug ..... i spose
anonymous
  • anonymous
?
amistre64
  • amistre64
s(2+h) -s(2) ----------- h

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More answers

amistre64
  • amistre64
its alot of work just to prove that derivatives work ....
anonymous
  • anonymous
lol true.
amistre64
  • amistre64
20-5(2+h)^2 -(20-5(2)^2) ------------------------ h 20-5(4 +4h +h^2) -20 +20 ------------------------ h 20-20-20h-5h^2 ---------------- h -20h-5h^2 ---------- = -20 -5h h
anonymous
  • anonymous
I think I have to use the formula (f(a)-f(x))/(a-x)
anonymous
  • anonymous
Can you explain to me how to use that formula?
amistre64
  • amistre64
now for the derivative we get: -10t ; at t=2 we get -20 when h=0, we get -20
amistre64
  • amistre64
its the same formula
anonymous
  • anonymous
oh no derivatives yet
anonymous
  • anonymous
can you explain to me how to use it? like what's a and what's x?
anonymous
  • anonymous
on the interval
amistre64
  • amistre64
yes, you are doing the proof for derivatives so that in the next few weeks your teacher will say, and this is how we do it the easy way...
anonymous
  • anonymous
haha ok.
amistre64
  • amistre64
f(a) and f(x) simply mean to use the formula with an a and an x instead of a t like this: f(a) - f(x) (20-5a^2) - (20-5x^2) ------------------- a-x
amistre64
  • amistre64
20-5a^2 -20+5x^2 ----------------- a - x -5a^2+5x^2 ------------- a - x -5(a^2-x^2) ------------- a - x -5(a+x)(a-x) ------------- a - x -5(a+x)
anonymous
  • anonymous
ok, what do you plug into f(a)?
anonymous
  • anonymous
in [2,2+h]
amistre64
  • amistre64
.............. a = 2 and x = h
anonymous
  • anonymous
?
anonymous
  • anonymous
is a=2 from the 2+h part or from just the 2 part.
amistre64
  • amistre64
it doesnt matter..... a =2 OR a=h, OR x=2 OR x=h.... -5(a+x) -5(2+h) is the same things as -5(h+2)
anonymous
  • anonymous
I'm confused.
anonymous
  • anonymous
Like, I know the forumla, I just don't understand how to use it.
amistre64
  • amistre64
then dont use it lol
amistre64
  • amistre64
does it matter if f(a) = 2 and f(x) = (2+h)? no
amistre64
  • amistre64
becasuse when h=0 it just disappears to -20
amistre64
  • amistre64
\[\frac{y-y_0}{x-x_0}\] do you understand this formula?
anonymous
  • anonymous
yes.
amistre64
  • amistre64
that is all it is; but they rename that parts ..... but it doesnt matter what they name them, they still do the same things
anonymous
  • anonymous
so the bottom part is a-x... don't you put 2-h?
anonymous
  • anonymous
why did you only put h?
amistre64
  • amistre64
\[\frac{f(a)-f(x)}{a-x} \iff \frac{y-y_0}{x-x_0} \iff \frac{f(x+h)-f(x)}{h}\]
anonymous
  • anonymous
oh ok.
amistre64
  • amistre64
because \((x+h)-(x)=h\)
anonymous
  • anonymous
why?
amistre64
  • amistre64
why what? why does x-x+h = 0+h = h?
anonymous
  • anonymous
oh ok. just kidding.
amistre64
  • amistre64
\[(x+h)-(x)\] \[x+h-x\] \[x-x+h\] \[(x-x)+h\] \[0+h\] \[h\]
anonymous
  • anonymous
so the formula would be \[(f(2+h)-f(2))\div(h)\]
amistre64
  • amistre64
yes
anonymous
  • anonymous
ok.
anonymous
  • anonymous
so its ((20-5(2+h)^2)-(20-5(2)^2))/h
amistre64
  • amistre64
yes
anonymous
  • anonymous
so (20-5(2+h)^2)/h
anonymous
  • anonymous
and finally -20-h?
amistre64
  • amistre64
yes
anonymous
  • anonymous
:) Thank you

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