Hi guys. equation of tangent line to the curve when x=1 i tried and got 0 can you show me how to solve with working please. thank you!
Last edited by tankertert; April 17th 2012 at 03:02 AM.
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dy/dx=6x-1 So when x=1 gradient of tangent=5 Also when x=1 y=2+e So you want a straight line with gradient 5 through the point (1,2+e)
Originally Posted by tankertert equation of tangent line to the curve when x=1 I suspect OP meant . If so OP remember to write it like this e^{1-x}.
Originally Posted by a tutor I suspect OP meant . If so OP remember to write it like this e^{1-x}. sorry yes that what i mena
So now dy/dx=6x-e^1-x When x=1 y=4 and dy/dx=5 Require straight line through (1,4) with gradient 5
Originally Posted by biffboy So now dy/dx=6x-e^1-x When x=1 y=4 and dy/dx=5 Require straight line through (1,4) with gradient 5 but how did you get that
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