What are the critical points of #f(x) =e^x-(xlnx)/(x-2)^2#?

Question

Christopher Agresta · Accepted Answer

See the answer below:

Axel Alvarez · Answer

undefined To find the critical points of $ f(x) = e^x - \frac{x \ln x}{(x-2)^2} $, we first need to find its derivative and then solve for $ f'(x) = 0 $.

First, find the derivative of $ f(x) $ using the product rule and chain rule:

$ f'(x) = e^x - \frac{d}{dx} \left( \frac{x \ln x}{(x-2)^2} \right) $

Using the quotient rule:

$ f'(x) = e^x - \frac{(x-2)^2 \frac{d}{dx}(x \ln x) - x \ln x \frac{d}{dx}(x-2)^2}{(x-2)^4} $

Now, compute the derivatives:
$ \frac{d}{dx}(x \ln x) = \ln x + 1 $
$ \frac{d}{dx}(x-2)^2 = 2(x-2) $

Substitute these derivatives back into $ f'(x) $:
$ f'(x) = e^x - \frac{(x-2)^2(\ln x + 1) - x \ln x \cdot 2(x-2)}{(x-2)^4} $

Simplify this expression:
$ f'(x) = e^x - \frac{(x-2)^2 \ln x + (x-2)^2 - 2x(x-2) \ln x}{(x-2)^4} $
$ f'(x) = e^x - \frac{(x-2)^2 \ln x + (x-2)^2 - 2x^2 + 4x \ln x}{(x-2)^4} $
$ f'(x) = e^x - \frac{x^2 \ln x - 4x \ln x + (x-2)^2 + 4x - 4}{(x-2)^4} $

Now, set $ f'(x) $ equal to zero and solve for $ x $:

$ e^x - \frac{x^2 \ln x - 4x \ln x + (x-2)^2 + 4x - 4}{(x-2)^4} = 0 $

Unfortunately, solving this equation analytically may not be straightforward. You may need to use numerical methods or approximation techniques to find the critical points of the function.