Consider the function h(x)=e^x. Determine the equation for the tangent line at x = 1. All values must be exact (no decimal values).

Question

i know the answer is just e but how do i prove it? by just subbing 1 into x? it cant be that easy..

Answer 1

h(x) = y = e^x
you know when x = 1, y = e
So the tangent makes contact at (1,e)
But you don't know the slope of the tangent!
dy/dx = e^x
when x = 1, dy/dx = e
So the slope of the tangent is e

tangent equation:
y -e = e(x - 1)
y = ex - e + e

y = ex is the equation of the tangent.

proof:
https://www.wolframalpha.com/input/?i=plot+y+%3D+e%5Ex,+y+%3D+ex

Answer 2

i don't understand how you got y-e=e(x-1)??

Answer 3

The point-slope forms straight line equation :

y − y1 = m ( x − x1 )

In this case x1 = 1

y1= eˣ¹ = e¹ = e

m = slope = first derivation in point x = 1

m = eˣ¹ = e¹ = e

y − y1 = m ( x − x1 )

y − e = e ( x − 1 )

y − e = e ∙ x − e ∙ 1

y − e = e ∙ x − e

add e to both sides

y − e + e = e ∙ x − e + e

y = e ∙ x