f(x) = (x-3)/(x+1) find f'(0) f(x) = (x-3)/(x+1) find f'(0)
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calendarEducator since 2011
write5,348 answers
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You need to find derivative of function using the quotient rule such that:
You need to substitute 0 for x in f'(x) such that:
Hence, evaluating the derivative of the function at yields
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calendarEducator since 2008
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f(x) = (x-3)/(x+1)
Let f(x) = u/v such that:\
u= x-3 ==> u' = 1
v= x+1 ==> v'= 1
f'(x) = (u'v- uv')/v^2
= [1*(x+1) - (x-3)*1]/ (x+1)^2
= (x+1 - x + 3)/ (x+1)^2
= 4/(x+1)^2
f'(0) = 4/(0+1)^2 = 4/1 = 4
==> f'(0) = 4
f(x) = (x-3)/(X+1) THEN
by putting 0 on the value of x we get,
f(0) = (0-3)/(0+1)
f(0) =-3/1
f(0) =-3
f(x) = (x-3) / (x+1)
As, x= 0
(plugging the values in the equation)
f(0) = (0-3) / (0+1)
f(0) = (-3) / (+1)
f(0) = -3
To calculate f'(0), we'll use the definition of the derivative:
f'(0) = lim {[f(x) - f(0)]/(x-0)} , x-->0
f'(0) = lim {[(x-3)/(x+1) - (-3)/1]/ x}, x-->0
f'(0) = lim {[(x-3)/(x+1) - (-3)/1]/ x}, x-->0
f'(0) = lim [(x-3 + 3x + 3)/ x(x+1)], x-->0
We'll combine like terms:
f'(0) = lim [(x-3 + 3x + 3)/ x(x+1)], x-->0
f'(0) = lim [4x/x(x+1)], x-->0
f'(0) = lim [4x/x(x+1)], x-->0
f'(0) = lim [4/(x+1)], x-->0
f'(0) = lim [4/(x+1)], x-->0
f'(0) = 4/(0+1) = 4/1 = 4
f(x) = (x-3)/(x+1)
To find f'(0).
Solution:
f(x) = (x+3)/(x+1) = [(x+1)- 4]/(x-1)
f(x) = (x+1)/(x+1) - 4/(x+1).
f(x) = 1 - 4/(x+1)
f'(x) = (1)' - {4/(x+1)}'
f'(x) = 0 - 4/(x+1)^2, as (constant)' = 0 and (k/(x+a)^n)' = - kn/(x-a)^(n+1).
Therefore, f'(0) = -4/(0+1)^2 = -4.
Given:
f(x) = (x - 3)/(x + 1)
Let:
x - 3 = u
And
x + 1 = v
Then:
f(x) = u/v
And:
f'(x) = (u'v - uv')/(v^2)
u' = 1
v' = 1
Substituting values of u, v, u' and v' in equation for f"(x):
f'(x) = [1*(x + 1) - 1*(x - 3)]/[(x + 1)^2]
= (x + 1 - x + 3)/[(x + 1)^2]
= 4/[(x + 1)^2]
Therefore:
f'(0) = 4/[(0 + 1)^2
= 4/1 = 4
Answer:
f'(0) = 4
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