Find the tangent line of $f(x)=e^{-x}\ln(x)$

Question

The function is $$f(x)=e^{-x}\ln(x)$$ and the point given is $(1,0)$

After differentiating we get $$f'(x)=-e^{-x}\ln(x)+\frac{e^{-x}}{x}$$ the problem I'm have is solving $f'(x)=0$, any ideas on how I should proceed?

Answer 1

What you have found is the slope function for your original function at any point where that function is defined. This includes the point that you're given. Just use $y-y_0=m(x-x_0)$ to form the equation of the tangent line. The point is given ($(x_0,y_0)=(1,0)$) and you know the slope at that point:

$$ y-0=f'(1)(x-1). $$

Answer 2

For the equation of the tangent line you need $f'(1)$ which is $e^{-1}$

That is your slope

Now use point slope formula and find the equation of the tangent line.

I found it to be $$y=e^{-1}(x-1)$$

Answer 3

The basis of the answer will be $y - y_{1} = m_{TAN}(x - x_{1}).$

Now, $(x_{1},y_{1}) = (1,0)$, and

$$ M_{TAN} = f'(1) = -e^{-1}(\ln1) + \frac{1}{e} = \frac{1}{e}.$$

Hence, $ y - 0 = \frac{1}{e}(x - 1) $; or rather,

$$ y = \frac{1}{e}(x - 1) = \frac{x}{e} - \frac{1}{e}.$$