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)
the slope for any x is
h'(x) = e^x
So, at the point (1,e) the slope is e
That means the equation of the line is
y-e = e(x-1)
See the graphs at
https://www.wolframalpha.com/input/?i=plot+y%3De%5Ex,+y%3Dex
To determine the equation for the tangent line at a specific point on a curve, we will need two pieces of information: the slope of the curve at that point and a point on the curve. In this case, we need to find the slope of the curve h(x) = e^x at the point x = 1.
To find the slope at x = 1, we can use the derivative of the function h(x). The derivative of h(x) = e^x with respect to x is also e^x. So, the derivative of h(x) is just e^x.
Now that we have the slope, we need a point on the curve. We know that we are interested in the tangent line at x = 1. So, we can find the corresponding y-coordinate by plugging x = 1 into the original function h(x) = e^x.
h(1) = e^1 = e
Therefore, the point on the curve is (1, e).
Now we have all the required information: the slope is e^1 = e, and the point on the curve is (1, e).
Using the point-slope form of a linear equation, the equation for the tangent line is given by:
y - y1 = m(x - x1)
where (x1, y1) is the point on the curve, and m is the slope.
Plugging in the values, we get:
y - e = e(x - 1)
Expanding the equation:
y - e = ex - e
Rearranging the terms:
ex - y = e - e
Simplifying:
ex - y = 0
Therefore, the equation for the tangent line to the curve h(x) = e^x at x = 1 is ex - y = 0.