1. ## tangent line curve

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!

2. ## Re: tangent line curve

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)

3. ## Re: tangent line curve

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}.

4. ## Re: tangent line curve

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

5. ## Re: tangent line curve

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

6. ## Re: tangent line curve

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