The equation of the line tangent to the curve at x = 11 is y = (5 / 432) · x - 43 / 432.
What is the equation of the line tangent to a given curve?
Herein we have a rational function, of which we must derive the equation of the line tangent to the curve at a given point. First, find the first derivative of the function by derivative rules:
f'(x) = [(6 · x + 6) - 6 · (x - 9)] / (6 · x + 6)²
f'(x) = 60 / (6 · x + 6)²
Second, evaluate the first derivative of the function:
f'(11) = 60 / (6 · 11 + 6)²
f'(11) = 5 / 432
This value represents the slope of the tangent line.
Third, calculate the y-value of the curve:
f(11) = (11 - 9) / (6 · 11 + 6)
f(11) = 1 / 36
Fourth, determine the intercept of the tangent line:
b = 1 / 36 - (5 / 432) · (11)
b = - 43 / 432
The equation of the line tangent to the curve at x = 11 is y = (5 / 432) · x - 43 / 432.
To learn more on tangent lines: https://brainly.com/question/17193273
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