Find the equation of the line tangent to the graph of f(x) = 11 − x + 4 ln (2x−1) at x = 1.
f'(x) = -1 + 2/(2x-1)
f(1) = 11-1+4+ln(1) = 14
f'(1) = -1 + 2/1 = 1
so, now you have a point and a slope, so the tangent line is
y-14 = 1(x-1)
To find the equation of the line tangent to the graph of the function f(x) at a specific point x = a, you need to determine both the slope and the point.
To find the slope of the tangent line, you need to take the derivative of the function f(x) and evaluate it at x = a. So, let's start by finding the derivative of f(x) = 11 − x + 4 ln(2x−1).
The derivative of f(x) with respect to x can be calculated using the sum and power rules of differentiation.
f'(x) = -1 + 4 * (1/(2x−1)) * 2
Simplifying this further, we have:
f'(x) = -1 + 8 / (2x−1)
Now, we need to find the value of the derivative at x = 1. We substitute x = 1 into the derivative expression:
f'(1) = -1 + 8 / (2 * 1 - 1)
= -1 + 8 / 1
= - 1 + 8
= 7
So, the slope of the tangent line to the graph of f(x) at x = 1 is 7.
Now that we have the slope, we need to find the y-coordinate of the point of tangency. We can do this by substituting x = 1 into the original function f(x):
f(1) = 11 − 1 + 4 ln(2 * 1 − 1)
= 10 + 4 ln(1)
= 10
Therefore, the point of tangency is (1, 10).
Now we can use the slope-intercept form of a linear equation to find the equation of the tangent line:
y - y1 = m(x - x1)
where (x1, y1) is the point of tangency, and m is the slope.
Plugging in the values, we get:
y - 10 = 7(x - 1)
Simplifying further:
y - 10 = 7x - 7
y = 7x + 3
Hence, the equation of the line tangent to the graph of f(x) = 11 − x + 4 ln(2x−1) at x = 1 is y = 7x + 3.