Respuesta :
Answer:
[tex]y=\dfrac{3x}{e}+\dfrac{4}{e}[/tex]
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).
[tex]g(x) =e^{x^3}[/tex]
- we know that [tex]\frac{d}{dx}(e^{f(x)})=e^{f(x)}f'(x)[/tex]
[tex]g'(x) =e^{x^{3}}(3 x^{2})[/tex]
[tex]g'(x) =3 x^{2} e^{x^{3}}[/tex]
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
[tex]m =3 (-1)^{2} e^{(-1)^{3}}\\m =3e^{-1}\\m=\dfrac{3}{e}[/tex]
using the equation of line:
[tex](y-y_1)=m(x-x_1)[/tex]
we'll find the equation of the tangent line.
here (x1,y1) =(-1,1/e), and m = 3/e
[tex](y-\dfrac{1}{e})=\dfrac{3}{e}(x+1)\\y=\dfrac{3x}{e}+\dfrac{3}{e}+\dfrac{1}{e}\\[/tex]
[tex]y=\dfrac{3x}{e}+\dfrac{4}{e}[/tex]
this is the equation of the tangent at point (-1,1/e)
