equation of tangent line to f(x)=(2x -5) / (x+1) at x=0
first i need to find the derivative and i got : 7/(x+1)^2
and then i need to plug 0 in? if so then it will equal 0 how would i
get the answer ; y= 7x -5 ?
first i need to find the derivative and i got : 7/(x+1)^2
and then i need to plug 0 in? if so then it will equal 0 how would i
get the answer ; y= 7x -5 ?
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f '(x) = ((x+1)(2) - (2x-5)) / (x+1)²
f '(x) = 7/(x+1)²
f '(0) = 7/(0+1)² = 7/1 = 7, not 0.
f(0) = -5 (plugging 0 into f(x))
y - (-5) = 7(x - 0)
y + 5 = 7x
y = 7x - 5
f '(x) = 7/(x+1)²
f '(0) = 7/(0+1)² = 7/1 = 7, not 0.
f(0) = -5 (plugging 0 into f(x))
y - (-5) = 7(x - 0)
y + 5 = 7x
y = 7x - 5
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YOUR DERIVATIVE IS QUITE RIGHT.
THE SLOPE OF THE TANGENT AT X = 0 IS 7/(X + 1) = 7/1 = 7
YOUR EQUATION IS Y = 7X + C
C IS THE CONSTANT WHEN X = 0 IN F(X) = (2X - 5)/( X + 1) I.E. 5
SO EQUATION IN Y = 7X -5
THE SLOPE OF THE TANGENT AT X = 0 IS 7/(X + 1) = 7/1 = 7
YOUR EQUATION IS Y = 7X + C
C IS THE CONSTANT WHEN X = 0 IN F(X) = (2X - 5)/( X + 1) I.E. 5
SO EQUATION IN Y = 7X -5
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f(x) = (2x - 5)/(x+1)
f(0) = -5
f ' (x) = [(x + 1)(2) - (2x - 5) ] /(x + 1)^2
f ' (0) = ( 2 + 5) /1 = 7
eqn of tangent is
y = f(0) + f '(0) (x - 0)
y = -5 + 7x
f(0) = -5
f ' (x) = [(x + 1)(2) - (2x - 5) ] /(x + 1)^2
f ' (0) = ( 2 + 5) /1 = 7
eqn of tangent is
y = f(0) + f '(0) (x - 0)
y = -5 + 7x