Which of the following best represents a portion of the graph $y = \frac{1}{e^x} + x - \frac{1}{e}$ near (1, 1)

Question

By taking derivatives, I know that the slope of this function is positive near (1, 1) and the magnitude of the slope will keep increasing. However, how can we tell the difference between (A) and (D)?

Thanks!

Answer 1

The difference between (A) and (D) is the apparent slope: the slope in (A) is greater than $1$, and the slope in (D) is between $0$ and $1$. That is, finding the sign of the slope at $x=1$ is not quite enough; you want to see how your result compares with $1$.

\begin{align} y&=e^{-x}+x-e^{-1}\\ y'&=-e^{-x}+1\\ y''&=e^{-x} \end{align}

Since $0\leq y'|_{x=1}\leq1$ and $y''|_{x=1}>0$, the correct answer appears to be (D).

Answer 2

D.

Note that $f(x)=e^{-x}+x+C$ is convex in $x$ for any constant $C$, and that the derivative of $f(x)$ is $1-e^{-x}$ which is $1-e < 1$ at $x=1$, actually it is about .6. So you want the graph where the function is convex and has a positive slope that is still less than 1.