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

Answer:

$y=\dfrac{3x}{e}+\dfrac{4}{e}$

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

Step-by-step 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 $\frac{d}{dx}(e^{f(x)})=e^{f(x)}f'(x)$

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

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

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

$m =3 (-1)^{2} e^{(-1)^{3}}\\m =3e^{-1}\\m=\dfrac{3}{e}$

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

$(y-\dfrac{1}{e})=\dfrac{3}{e}(x+1)\\y=\dfrac{3x}{e}+\dfrac{3}{e}+\dfrac{1}{e}$

$y=\dfrac{3x}{e}+\dfrac{4}{e}$

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