Question

What is the equation of the tangent line of `f(x) =e^(3x-1)-(3x-1)` at `x=2`?

Answer 1

`0.60x-0.00135y-1=0` is quite close to the tangent at P(2, -143.4)#

Explanation:

The point of contact of the tangent P is `(2, e^5-5)=(2, 143.4)`.

The slope of the tangent is

`f'= 3e^(3x-1)-3=442.2`, at P.

So, the equation to the tangent at P is

`y-143.4=442.2(x-2)`, giving

`0.60x-0.00135y-1=0`

The first graph relative positions of tangent and the curve, near the

x-axis. The point of contact is in hiding.

The second graph graph is in the proximity of the Point of contact that is also marked, at y = 143.4.

graph{(e^(3x-1)-3x+1-y)(.6x-0.00135y-1)=0 [-10, 10, -5, 5]}
graph{(e^(3x-1)-3x+1-y)(.6x-0.00135y-1)((x-2)^2+(y-143.4)^2-.004)=0 [0, 4, 137, 147]}