Final answer:
The equation of the tangent line to the function y = ln(x² - 8) at the point (3, 0) is y = 6x - 18.
Step-by-step explanation:
To find an equation of the tangent line to the curve at the point (3, 0) for the function y = ln(x² - 8), we first need to calculate the derivative of the function, which gives us the slope of the tangent line at any point.
The derivative of y with respect to x is:
dy/dx = 1/(x² - 8) × 2x
At x = 3, the derivative is:
dy/dx = 2 × 3 / (3² - 8) = 6 / 1 = 6
This means the slope of the tangent line at (3, 0) is 6. Now, we use the point-slope form of the equation of a line, which is y - y1 = m(x - x1), where m is the slope and (x1, y1) is a point on the line. Since the slope m=6 and the point is (3, 0), the equation of the tangent line is:
y - 0 = 6(x - 3)
Simplifying, we get