Consider the function h(x)=ex. Determine the equation for the tangent line at x = 1. All values must be exact (no decimal values)
To determine the equation for the tangent line at x = 1, we need to find the slope and point of the tangent line.
Step 1: Find the slope of the tangent line
The slope of the tangent line at a given point can be found by taking the derivative of the function. For h(x) = ex, the derivative is given by:
h'(x) = d/dx(ex) = ex
Step 2: Evaluate the derivative at x = 1
To find the slope of the tangent line at x = 1, we need to evaluate the derivative at x = 1:
h'(1) = e^(1) = e
Step 3: Determine the point on the function h(x) at x = 1
Evaluate the function h(x) = ex at x = 1:
h(1) = e^1 = e
Step 4: Write the equation of the tangent line
Using the point-slope form of a line, the equation of the tangent line is given by:
y - y1 = m(x - x1)
Substituting the values we found:
y - e = e(x - 1)
Rearranging the equation:
y - e = ex - e
Simplifying the equation gives us the final answer:
y = ex + e - e
Therefore, the equation for the tangent line at x = 1 for the function h(x) = ex is:
y = ex + 1
To determine the equation for the tangent line of the function h(x) = ex at x = 1, we need to find the derivative of h(x) first.
The derivative of h(x) with respect to x represents the slope of the tangent line at any given point x. In this case, we want to find the derivative at x = 1.
The derivative of ex is simply ex. So, the derivative of h(x) is h'(x) = ex.
To find the slope of the tangent line at x = 1, we substitute x = 1 into the derivative:
h'(1) = e(1) = e
Now, we have the slope of the tangent line which is e. To find the equation of the tangent line, we also need a point on the line. In this case, the point is (1, h(1)).
To find h(1), we substitute x = 1 into the original function h(x):
h(1) = e(1) = e
Therefore, the point on the tangent line is (1, e).
Using the point-slope form of the equation of a line, which is y - y1 = m(x - x1), where m represents the slope and (x1, y1) represents a point on the line, we substitute the values we found:
y - e = e(x - 1)
Simplifying the equation:
y - e = ex - e
Finally, rearranging the equation to the standard form:
ex - y = e - e
ex - y = 0
So, the equation for the tangent line of h(x) = ex at x = 1 is ex - y = 0.