Determine an equation of the tangent line to the curve of f (x) = 3e^4x − 3 at the point where the curve crosses the y-axis.
y=0 when x=0, so we are looking at the line through (0,0)
f'(x) = 12e^4x
f'(0) = 12
the line is thus y=12x
The curve crosses the y axis when x = 0.
f(0) = 3*e^0 - 3 = 0 @ x=0
dy/dx = 12 e^4x = 12 @ x=0
The straight tangent line at the origin (the tanglent point) is
y = 12 x.
To determine the equation of the tangent line to a curve at a given point, we need to find both the slope of the tangent line and the coordinates of the point of tangency.
In this case, we are looking for the slope of the tangent line at the point where the curve crosses the y-axis, which means that the x-coordinate of the point of tangency is 0. To find the y-coordinate of the point of tangency, we need to evaluate the function f(x) at x = 0.
Let's start by finding the y-coordinate of the point of tangency:
f(x) = 3e^(4x) - 3
When the curve crosses the y-axis, x = 0:
f(0) = 3e^(4*0) - 3
f(0) = 3e^0 - 3
f(0) = 3(1) - 3
f(0) = 3 - 3
f(0) = 0
So the y-coordinate of the point of tangency is 0.
Now, let's find the derivative of the function f(x) using the exponential function rule:
f'(x) = d/dx (3e^(4x) - 3)
= 12e^(4x)
To find the slope of the tangent line at x = 0, we evaluate the derivative at that point:
f'(0) = 12e^(4*0)
f'(0) = 12e^0
f'(0) = 12(1)
f'(0) = 12
Now we have the slope of the tangent line at x = 0, which is 12. We also have the coordinates of the point of tangency, which are (0, 0).
The equation of a line can be written as y = mx + b, where m is the slope and b is the y-intercept. In this case, since the line crosses the y-axis at (0, 0), the y-intercept is 0.
So, the equation of the tangent line to the curve f(x) = 3e^(4x) - 3 at the point where the curve crosses the y-axis is:
y = 12x + 0
which simplifies to:
y = 12x