If the function f is defined by f(x)=x^2-2x, find limit h->0
(f(3+h)-(f(3-h))/2h
first, f(3+h)=(3+h)^2 -2(3+h)
= 9+6h+h^2-6-2h = h^2+4h+3= (h+3)(h+1)
second f(3-h)=(3-h)^2 -2(3-h)=
= 9-6h+h^2-6+2h=h^2-4h+3=(h-3)(h-1)
so first -second then is
h^2-h^2+4h+4h +3-3=8h
finally, lim 8h/2h=4
check all that.
Thank you for your help!
For the second part I thought it would be -(9+6h-h^2)-6+2h?
you put a minus sign in front of the first term? Why?
and, the 6h term in (3-h)^2=9-6h+h^2
(f(3+h)-(f(3-h))/2h
I put a minus sign because it has a negative in front of the f(3-h) so it's -(f(3-h))
To find the limit as h approaches 0 of the given function, let's start by substituting f(3+h) and f(3-h) into the expression and simplify it.
f(x) is defined as f(x) = x^2 - 2x
Substituting f(3+h):
f(3+h) = (3+h)^2 - 2(3+h)
= (9 + 6h + h^2) - (6 + 2h)
= 9 + 6h + h^2 - 6 - 2h
= 3 + 4h + h^2
Substituting f(3-h):
f(3-h) = (3-h)^2 - 2(3-h)
= (9 - 6h + h^2) - (6 - 2h)
= 9 - 6h + h^2 - 6 + 2h
= 3 - 4h + h^2
Now, substitute these values back into the original expression:
[(3 + 4h + h^2) - (3 - 4h + h^2)] / 2h
Simplifying further:
[(3 + 4h + h^2 - 3 + 4h - h^2)] / 2h
= (8h) / 2h
= 4
Therefore, the limit of the given function as h approaches 0 is 4.