Question 418868
{{{f(x) = log(6, (5x + x^(-5)))}}}
First let's find the y-coordinate of "the point indicated":
{{{f(1) = log(6, (5(1) + (1)^(-5)))}}}
{{{f(1) = log(6, (5 + 1))}}}
{{{f(1) = log(6, (6))}}}
f(1) = 1
So "the point indicated" is (1, 1)<br>
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) = {{{(1/ln(6))(1/(5x + x^(-5)))(5 + (-5)x^(-6))}}}
To find the slope at (1, 1) we need
f'(1) = {{{(1/ln(6))(1/(5(1) + (1)^(-5)))(5 + (-5)(1)^(-6))}}}
which simplifies as follows:
f'(1) = {{{(1/ln(6))(1/(5 + 1))(5 + (-5))}}}
f'(1) = {{{(1/ln(6))(1/6)(0)}}}
f'(1) = 0<br>
A line, with a slope of 1, though the point (1, 1) would be y = 1. This is the equation for the tangent line.<br>
Here's a graph of f(x):
{{{graph(400, 400, 0.1, 3, -5, 10, ln(5x + x^(-5))/ln(6))}}}