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derivative

The equation of the tangent to the curve 1/(1−2x) at the point (1, -1) is given by y=ax+b, where
a=
b=
(edited 6 years ago)
Original post by fahd20428
The equation of the tangent to the curve 11−2x at the point (1, -1) is given by y=ax+b, where
a=
b=


y=11-2x is a straight line. The tangent to a straight line is the line itself. The line doesn't pass (1,-1), have you written the question correctly?
(edited 6 years ago)
Reply 2
Original post by NotNotBatman
y=11-2x is a straight line. The tangent to a straight line is the line itself. The line doesn't pass (1,-1), have you written the question correctly?


sorry it is 1/(1-2x)
Original post by fahd20428
sorry it is 1/(1-2x)


Find the derivative of 1/(1-2x), the derivative is the gradient function, which will allow you to find out the gradient of the tangent at some point x. sub x=1 into the gradient function and use the formula y-y1=m(x-x1)
Reply 4
Original post by NotNotBatman
Find the derivative of 1/(1-2x), the derivative is the gradient function, which will allow you to find out the gradient of the tangent at some point x. sub x=1 into the gradient function and use the formula y-y1=m(x-x1)


is the answer going to be a=-2 b=1

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