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]#

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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Answer 2

To find the equation of the tangent line of the function f(x) = x/(x-2e^x) + x/e^x - x at x = 3, we need to find the derivative of the function and evaluate it at x = 3.

First, let's find the derivative of f(x).

f'(x) = (d/dx)[x/(x-2e^x)] + (d/dx)[x/e^x] - (d/dx)[x]

Using the quotient rule and the chain rule, we can simplify this expression:

f'(x) = [(1*(x-2e^x) - x*(1-2e^x))/(x-2e^x)^2] + [(1e^x - x(e^x))/(e^x)^2] - 1

Simplifying further:

f'(x) = [(x - 2e^x - x + 2xe^x)/(x-2e^x)^2] + [(e^x - xe^x)/(e^x)^2] - 1

f'(x) = [(2xe^x - e^x)/(x-2e^x)^2] + [(e^x - xe^x)/(e^x)^2] - 1

Now, let's evaluate f'(x) at x = 3:

f'(3) = [(2(3)e^3 - e^3)/(3-2e^3)^2] + [(e^3 - 3e^3)/(e^3)^2] - 1

Simplifying further:

f'(3) = [(6e^3 - e^3)/(3-2e^3)^2] + [(-2e^3)/(e^6)] - 1

f'(3) = [(5e^3)/(3-2e^3)^2] - [2e^3/(e^6)] - 1

Now, we have the slope of the tangent line at x = 3. To find the equation of the tangent line, we use the point-slope form:

y - y1 = m(x - x1)

Using the point (3, f(3)) and the slope f'(3), we can substitute the values into the equation:

y - f(3) = f'(3)(x - 3)

Simplifying further:

y - [f(3)] = f'(3)(x - 3)

This is the equation of the tangent line of f(x) = x/(x-2e^x) + x/e^x - x at x = 3.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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