Question

57 - 62. transcendental functions determine the end behavior of the following transcendental functions by analyzing appropriate limits. then provide a simple sketch of the associated graph, showing asymptotes if they exist. 57. $f(x)=-3e^{-x}$ 58. $f(x)=2^{x}$ 59. $f(x)=1 - ln x$

Explanation:

Step1: Analyze $\lim_{x

ightarrow+\infty}f(x)$ for $f(x)= - 3e^{-x}$
We know that the exponential - function property: $\lim_{x
ightarrow+\infty}e^{-x}=\lim_{x
ightarrow+\infty}\frac{1}{e^{x}} = 0$. So, $\lim_{x
ightarrow+\infty}-3e^{-x}=-3\lim_{x
ightarrow+\infty}e^{-x}=0$.

Step2: Analyze $\lim_{x

ightarrow-\infty}f(x)$ for $f(x)= - 3e^{-x}$
As $x
ightarrow-\infty$, $-x
ightarrow+\infty$. Then $\lim_{x
ightarrow-\infty}-3e^{-x}=-\infty$ since $e^{-x}$ grows without bound as $x
ightarrow-\infty$.

Step3: Sketch the graph

The function $y = - 3e^{-x}$ has a horizontal asymptote $y = 0$ as $x
ightarrow+\infty$. The function is always negative, and it decreases from $-\infty$ (as $x
ightarrow-\infty$) towards the $x$ - axis (as $x
ightarrow+\infty$).

Answer:

As $x
ightarrow+\infty$, $f(x)
ightarrow0$; as $x
ightarrow-\infty$, $f(x)
ightarrow-\infty$. The graph has a horizontal asymptote $y = 0$ as $x
ightarrow+\infty$.