SOLUTION: Find an equation of the tangent line at the point indicated. f (x) = log3 (8x + x^-8), x = 1?
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Question 352439: Find an equation of the tangent line at the point indicated. f (x) = log3 (8x + x^-8), x = 1?
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
f (x) =
f(x) = (1/ln(3))*ln(8x + x^-8)
f'(x) = (1/ln(3))*(1/(8x + x^-8))*(8 - 8x^-9)
@ x = 1: f'(x) = (1/ln(3))*(1/(8 + 1))*(8 - 8)
= 0
---------
f(1) = log(3,9) = 2
--> y = 2 is the line tangent at x = 1
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