Answer: y = (-28)x + 14
Explanation:
The equation for the line tangent at point (a, b) can be found using the formula y = mx + c, where m is the slope of the line and c is the y-intercept. At point (6, 14), we need to find the value of m.
We start by finding the derivative of the original function g(x). Its derivative is given by:
dg(x)/dx=7/(x^2-3x+1)
Then, substitute x = 6 into the derivative expression to obtain:
dg(6)/dx = d/dx [7 * ln|x-3| ] evaluated at x = 6 = 7/3
Next, evaluate the original function g(x) at x = 6 to get g(6) = 7 * ln |6 - 3| / (6 - 3) = 7 * ln 3.
Since we know the coordinates of the point of tangency (6, 14), we can substitute them into the general form of the linear equation y = mx + c:
14 = 7 * 6 + c
14 = 42 + c
c = -28
The final equation of the line tangent at point (6, 14) is therefore:
y = (-28)x + 14