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).
i know the answer is just e but how do i prove it? by just subbing 1 into x? it cant be that easy..
i don't understand how you got y-e=e(x-1)??
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
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
To determine the equation for the tangent line at a given point on a curve, we need to find both the slope and the point of tangency.
Let's start by finding the slope of the tangent line at x = 1. The slope of the tangent line at a point on the curve represented by the function h(x) can be found using the derivative of h(x).
The function h(x) = e^x can be differentiated using the power rule of differentiation. The derivative of e^x with respect to x is simply e^x itself.
Therefore, the derivative of h(x) = e^x is h'(x) = e^x.
Now, we can find the slope of the tangent line at x = 1 by substituting x = 1 into the derivative h'(x):
h'(1) = e^1 = e.
So, the slope of the tangent line at x = 1 is e.
Next, we need to find the point of tangency. To do this, we substitute x = 1 into the original function h(x) = e^x:
h(1) = e^1 = e.
Therefore, the point of tangency is (1, e).
Now that we have the slope (e) and a point (1, e), we can use the point-slope form of a linear equation to find the equation of the tangent line:
y - y₁ = m(x - x₁),
where (x₁, y₁) represents the point of tangency and m represents the slope.
Substituting the values into the equation, we get:
y - e = e(x - 1).
Expanding the right side:
y - e = ex - e.
Rearranging the equation:
y = ex - e + e.
Simplifying:
y = ex.
Therefore, the equation for the tangent line at x = 1 is y = ex.