Question

What is the equation of the tangent line of `f(x) =x/(x-2e^x)+x/e^x-x` at `x=3`?

Answer 1

Not simplified *
Averaged to 4 decimal places
`[y=-1.0414(x-3)-2.9313]`

Explanation:

We identify that this function requires the use of the quotient rule multiple times to find the equation for the tangent line.

first find the derivative of each piece of the function separately.
`d/dx ((x)/(x-2e^x)) = ((x-2e^x)-x(1-2e^x))/((x-2e^x))^2`
simplifying this function we find `(2xe^x-2e^x)/((x-2e^x))^2`

*Now the second part of the function `d/dx((x)/(e^x))` Using the quotient rule `(e^x-xe^x)/((e^x))^2` which `(e^x)^2= e^(2x)`

`d/dx -x=-1`

From the rules of limits for derivatives, we come to the conclusion that `d/dx ((x)/(x-2e^x)) + ((x)/(e^x)) - x = (2xe^x-2e^x)/((x-2e^x))^2 + (e^x-xe^x)/((e^x))^2 - 1`
from what we know of what a derivative is we come to the consensus that plugging in a value of `f'(x) = m`
Also we know slope formula for a function is `y-y1=m(x-x1)`
using our original function we find a value of `f(3) ~~ -2.9313`
`y+2.93131=-1.0414(x-3)` which putting into slope intercept form looks like `y=-1.0414(x-3)-2.93131` This is of course simplified

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