SOLUTION: Find an equation of the tangent line at the point indicated. f (x) = log6 (5x + x^-5), x = 1

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Find an equation of the tangent line at the point indicated. f (x) = log6 (5x + x^-5), x = 1       Log On


   

Question 418868: Find an equation of the tangent line at the point indicated.
f (x) = log6 (5x + x^-5), x = 1
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
f%28x%29+=+log%286%2C+%285x+%2B+x%5E%28-5%29%29%29
First let's find the y-coordinate of "the point indicated":
f%281%29+=+log%286%2C+%285%281%29+%2B+%281%29%5E%28-5%29%29%29
f%281%29+=+log%286%2C+%285+%2B+1%29%29
f%281%29+=+log%286%2C+%286%29%29
f(1) = 1
So "the point indicated" is (1, 1)

Now we find the slope of the tangent at this point. For this we will need the first derivative Using the dervative of a logarithm plus the chain rule we get:
f'(x) = %281%2Fln%286%29%29%281%2F%285x+%2B+x%5E%28-5%29%29%29%285+%2B+%28-5%29x%5E%28-6%29%29
To find the slope at (1, 1) we need
f'(1) =
which simplifies as follows:
f'(1) = %281%2Fln%286%29%29%281%2F%285+%2B+1%29%29%285+%2B+%28-5%29%29
f'(1) = %281%2Fln%286%29%29%281%2F6%29%280%29
f'(1) = 0

A line, with a slope of 1, though the point (1, 1) would be y = 1. This is the equation for the tangent line.

Here's a graph of f(x):
graph%28400%2C+400%2C+0.1%2C+3%2C+-5%2C+10%2C+ln%285x+%2B+x%5E%28-5%29%29%2Fln%286%29%29