Calculus: Early Transcendentals 8th Edition

At $x=\ln2$, the graph of $f(x)$ has a horizontal tangent.
$$f(x)=e^x-2x$$ 1) First, find the derivative of function $f(x)$ $$f'(x)=\frac{d}{dx}(e^x)-2\frac{d}{dx}(x)$$ $$f'(x)=e^x-2$$ 2) The slope of the tangent line of the curve $f(x)$ at a point $A (x,y)$ would be the value of $f'(x)$. A horizontal tangent line is one that has a slope value of $0$, or, in other words, $f'(x)=0$ Therefore, to find at what value of $x$, the graph of $f(x)$ has a horizontal tangent, we need to solve $$f'(x)=0$$$$e^x-2=0$$$$e^x=2$$$$x=\ln2$$ That means at $x=\ln2$, the graph of $f(x)$ has a horizontal tangent.