# f(x) = (x-3)/(x+1)     find f'(0)f(x) = (x-3)/(x+1)     find f'(0)

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

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

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

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

khansaqib | Student, Grade 9 | (Level 1) Honors

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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

irza | Student, Grade 10 | (Level 1) eNoter

Posted on

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

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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

neela | High School Teacher | (Level 3) Valedictorian

Posted on

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.

krishna-agrawala | College Teacher | (Level 3) Valedictorian

Posted on

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