Chapter 3 - Section 3.1 - Derivatives of Polynomials and Exponential Functions - 3.1 Exercises - Page 183: 60

$x=\ln 2$

$f(x)=e^x-2x$ Recall: $f(x)$ has a horizontal tangent at values of $x$ for which $f'(x)=0$. Find $f'(x)$: $f'(x)=\frac{d}{dx}(e^x-2x)$ (Use the difference rule) $f'(x)=\frac{d}{dx}(e^x)-\frac{d}{dx}(2x)$ $f'(x)=e^x-2$ Find $x$ such that $f'(x)=0$: $e^x-2=0$ $e^x=2$ $x=\ln 2$. Thus, $f(x)$ has a horizontal tangent for $x=\ln 2$.