Question

Finding an Equation of a tangent Line in Exercise, find an equation of the tangent line to the graph of the function at the given point.

g(x) = e^x3 , (-1,1/e)

Answer 1

`y=(3x)/(e)+(4)/(e)`

this is the equation of the tangent at point (-1,1/e)

Explanation:

to find the tangent line we need to find the derivative of the function g(x).

`g(x) =e^(x^3)`

• we know that
  `(d)/(dx)(e^(f(x)))=e^(f(x))f'(x)`

`g'(x) =e^{x^(3)}(3 x^(2))`

`g'(x) =3 x^(2) e^{x^(3)}`

this the equation of the slope of the curve at any point x and it also the slope of the tangent at any point x. hence, g'(x) can be denoted as 'm'

to find the slope at (-1,1/e) we'll use the x-coordinate of the point i.e. x = -1

`m =3 (-1)^(2) e^{(-1)^(3)}\\m =3e^(-1)\\m=(3)/(e)`

using the equation of line:

`(y-y_1)=m(x-x_1)`

we'll find the equation of the tangent line.

here (x1,y1) =(-1,1/e), and m = 3/e

`(y-(1)/(e))=(3)/(e)(x+1)\\y=(3x)/(e)+(3)/(e)+(1)/(e)\\`

`y=(3x)/(e)+(4)/(e)`

this is the equation of the tangent at point (-1,1/e)