Question

(a) use a graphing device to graph f(x) = e^x - 2x.

Explanation:

Step1: Analyze function behavior

The function $f(x)=e^{x}-2x$. The derivative $f^\prime(x)=e^{x}-2$. Set $f^\prime(x) = 0$, then $e^{x}-2=0$, so $x = \ln(2)\approx0.693$. When $x<\ln(2)$, $f^\prime(x)<0$, the function is decreasing. When $x>\ln(2)$, $f^\prime(x)>0$, the function is increasing. Also, $f(0)=e^{0}-2\times0 = 1$, and as $x\to-\infty$, $e^{x}\to0$ and $f(x)\to+\infty$, as $x\to+\infty$, $e^{x}$ grows much faster than $2x$ so $f(x)\to+\infty$.

Step2: Match with graphs

Based on the above - analysis of increasing - decreasing behavior and end - behavior, we can match the correct graph.

Answer:

The graph that has a minimum point at approximately $x = \ln(2)\approx0.693$, is decreasing for $x<\ln(2)$ and increasing for $x>\ln(2)$, and has $y - intercept$ at $(0,1)$ and goes to $+\infty$ as $x\to\pm\infty$. Without specific labels on the given graphs, it's not possible to point out a particular one by number or position, but the described characteristics should be used to identify the correct graph among the options.