Question

What is the slope of `f(x)=-xe^x/x^2` at `x=1`?

Answer 1

Slope of `f(x)` at `x = 1` is `0`

Explanation:

`f(x) = -x * e^x/x^2` `= -e^x/x` = `-e^x*x^-1`

`f'(x) = -(e^x*(-x^-2) + x^-1*e^x )` (Product, Exponential and Power Rules)

The slope of `f(x)` at `x=1` is given by `f'(1)`

`f'(1) = -(e^1 * (-1) + e^1 *1)`

`f'(1) = -(-e+e) = 0`

Hence: Slope of `f(x)` at `x = 1` is `0`