Given: f(x) = (x-2) / (x+1) find the tangent line at x = 1.
I'm having trouble with this one. Been working on it for a bit, but feel I haven't gotten anywhere. Any help or explanation on how to do this problem would be much appreciated.
I'm having trouble with this one. Been working on it for a bit, but feel I haven't gotten anywhere. Any help or explanation on how to do this problem would be much appreciated.
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f(x) = (x-2) / (x+1)
f ' (x) = (x+1) - (x-2) / (x+1)^2
f ' (x) = 3 / (x+1)^2
@ x = 1
f ' (x) = 3/4 <== Slope
y = 3/4x + b
@ x = 1, initial equation is -1/2
Answer: y = 3/4x - 5/4
f ' (x) = (x+1) - (x-2) / (x+1)^2
f ' (x) = 3 / (x+1)^2
@ x = 1
f ' (x) = 3/4 <== Slope
y = 3/4x + b
@ x = 1, initial equation is -1/2
Answer: y = 3/4x - 5/4
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In order to find an equation for a line, you must find a point on the line and the slope. One is the x value of the point, so then plug one in for x in the function in order to find the y value. Since the derivative of a function is a function's slope, you can find the derivative. Use the equation y-y=m(x-x). I Hope this helps. Add details if this is not enough. :)
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Do you know Calculus? f(1) = -1/2
Find f'(1). Implicit differentiation gives you dy/dx = (1-y)/(x+1). Plug in and you get dy/dx = 3/4. Then you follow the formula for the tangent line y= (3/4) (x-1) + f(1). Once you simplify, you get y=3/4 x - 5/4.
Find f'(1). Implicit differentiation gives you dy/dx = (1-y)/(x+1). Plug in and you get dy/dx = 3/4. Then you follow the formula for the tangent line y= (3/4) (x-1) + f(1). Once you simplify, you get y=3/4 x - 5/4.