57 - 62. transcendental functions determine the end behavior of the following transcendental functions by…

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.\n57. $f(x)=-3e^{-x}$ 58. $f(x)=2^{x}$ 59. $f(x)=1 - ln x$\n60. $f(x)=|ln x|$ 61. $f(x)=sin x$ 62. $f(x)=\frac{50}{e^{2x}}$

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.\n57. $f(x)=-3e^{-x}$ 58. $f(x)=2^{x}$ 59. $f(x)=1 - ln x$\n60. $f(x)=|ln x|$ 61. $f(x)=sin x$ 62. $f(x)=\frac{50}{e^{2x}}$

Answer

Explanation:

Step1: Analyze $\lim_{x\rightarrow+\infty}-3e^{-x}$

We know that the exponential - function $y = e^{-x}=\frac{1}{e^{x}}$. As $x\rightarrow+\infty$, $e^{x}\rightarrow+\infty$, so $\lim_{x\rightarrow+\infty}-3e^{-x}=\lim_{x\rightarrow+\infty}-\frac{3}{e^{x}} = 0$.

Step2: Analyze $\lim_{x\rightarrow-\infty}-3e^{-x}$

As $x\rightarrow-\infty$, $-x\rightarrow+\infty$, and $e^{-x}\rightarrow+\infty$. So $\lim_{x\rightarrow-\infty}-3e^{-x}=-\infty$.

Step3: Determine asymptotes

Since $\lim_{x\rightarrow+\infty}-3e^{-x}=0$, the horizontal asymptote is $y = 0$ as $x\rightarrow+\infty$. There are no vertical asymptotes for the function $y=-3e^{-x}$ because the domain of $y = - 3e^{-x}$ is $(-\infty,+\infty)$.

To sketch the graph:

  • The function $y=-3e^{-x}$ is always negative because $e^{-x}>0$ for all real - valued $x$ and the coefficient $-3$ is negative.
  • As $x$ increases, the function approaches the $x$ - axis ($y = 0$) from below. As $x$ decreases, the function decreases without bound.

Answer:

As $x\rightarrow+\infty$, $f(x)\rightarrow0$; as $x\rightarrow-\infty$, $f(x)\rightarrow-\infty$. The horizontal asymptote is $y = 0$ as $x\rightarrow+\infty$.