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

Sep 30, 2016

$2 \left(1 - e\right) \cong - 3.43656$

#### Explanation:

$f \left(x\right) = - x {e}^{x} + 2 x$

$f ' \left(x\right) = - x {e}^{x} + \left(- 1\right) {e}^{x} + 2$ Product rule and power rule

$= - {e}^{x} \left(x + 1\right) + 2$

The slope of $f \left(x\right)$ at $x = 1$ is given by $f ' \left(1\right)$

$f ' \left(1\right) = - 2e+2 = 2 \left(1 - e\right)$

$f ' \left(1\right) \cong - 3.43656$

This can be seen by the graph of $f \left(x\right)$ in the region of $x = 1$ below:

graph{-xe^x+2x [-0.202, 2.498, -1.007, 0.343]}

Sep 30, 2016

Slope at $x = 1$ is $- 3.4365$

#### Explanation:

Given -

$y = - x {e}^{x} + 2 x$

Its slope is defined by its first derivative.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left[- x {e}^{x} \left(1\right) + {e}^{x} \left(- 1\right)\right] + 2$
$\frac{\mathrm{dy}}{\mathrm{dx}} = - x {e}^{x} - {e}^{x} + 2$

Its slope at $x = 1$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \left(1\right) {e}^{1} - {e}^{1} + 2$
$\frac{\mathrm{dy}}{\mathrm{dx}} = - {e}^{1} - {e}^{1} + 2 = - 3.4365$