Hi guys. equation of tangent line to the curve $\displaystyle y = 3x^2 + e^{1-x}$ when x=1 i tried and got 0 can you show me how to solve with working please. thank you!
Last edited by tankertert; Apr 17th 2012 at 02:02 AM.
Follow Math Help Forum on Facebook and Google+
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 $\displaystyle y = 3x^2 + e^1-x$ when x=1 I suspect OP meant $\displaystyle y = 3x^2 + e^{1-x}$. If so OP remember to write it like this e^{1-x}.
Originally Posted by a tutor I suspect OP meant $\displaystyle y = 3x^2 + e^{1-x}$. 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
View Tag Cloud