## University Calculus: Early Transcendentals (3rd Edition)

There is one point $(0,-1)$ where the slope of the curve is $2$.
$$y=x-e^{-x}$$ 1) First, find the derivative of $y$: $$y'=1-e^{-x}(-x)'$$ $$y'=1-e^{-x}(-1)$$ $$y'=1+e^{-x}$$ 2) To find if there are any points where the slope is $2$, we find if there are any values of $x$ so that $y'=2$ or not. $$y'=2$$ $$1+e^{-x}=2$$ $$e^{-x}=1=e^0$$ $$-x=0$$ $$x=0$$ - For $x=0$: $$y=0-e^{-0}=-e^0=-1$$ So there is one point $(0,-1)$ where the slope of the curve is $2$.