What is the slope of $f (x) = - x \frac{e^{x}}{x^{2}}$ $f(x)=-xe^x/x^2$ at $x = 1$ $x=1$ ?

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Answer 1 · 2016-05-11T02:06:01

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

$f (x) = - x \cdot \frac{e^{x}}{x^{2}}$ $f(x) = -x * e^x/x^2$ $= - \frac{e^{x}}{x}$ $= -e^x/x$ = $- e^{x} \cdot x^{- 1}$ $-e^x*x^-1$

$f'(x) = -(e^x*(-x^-2) + x^-1*e^x )$ (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 $0$

What is the slope of f(x)=-xe^x/x^2f(x)=−xexx2f(x)=-xe^x/x^2 at x=1x=1x=1?