Precalculus 2014

 PAGE: 166 SET: Exercises PROBLEM: 1

The function is $f(x)=2^{-x}$.

Make the table of values to find ordered pairs that satisfy the function.

Choose values for $x$ and find the corresponding values for $y$

 $x$ $f(x)=2^{-x}$ $(x, y)$ $-4$ $f(-4)=2^{-(-4)}=16$ $(-4, 16)$ $-3$ $f(-3)=2^{-(-3)}=8$ $(-3, 8)$ $-2$ $f(-2)=2^{-(-2)}=4$ $(-2, 4)$ $0$ $f(0)=2^{0}=1$ $(0, 1)$ $2$ $f(2)=2^{-(2)}=0.25$ $(2, 0.25)$ $4$ $f(4)=2^{-(4)}=0.06$ $(4, 0.06)$ $6$ $f(6)=2^{-(6)}=0.01$ $(6, 0.01)$ $8$ $f(8)=2^{-(8)}=0.003$ $(8, 0.003)$

Graph:

1. Draw a coordinate plane.

2. Plot the coordinate points.

3. Then sketch the graph, connecting the points with a smooth curve.

Observe the above graph :

Domain of the function is all real numbers.

Range of the function is $(0, \infty )$.

The function does not have the x-intercept.

$y$ - intercepts is $1$.

The line $y=L.$ is the horizontal asymptote of the curve $y=f(x)$.

if either $\lim_{x \to \infty}f(x)=L$ or $\lim_{x \to -\infty}f(x)=L$.

$\lim_{x \to \infty}2^{-x}$

$\\\lim_{x \to \infty}\frac{1}{2^{x}}\\\\=\frac{1}{2^{\infty}}\\\\$

$=0$.

$y=0$ is the horizontal asymptote of the function.

Observe the above graph the function does not have vertical asymptote.

End behavior : $\lim_{x\rightarrow -\infty }(f(x))=\infty$ and $\lim_{x\rightarrow \infty }f(x)=0$.

The function is continuous for all real numbers.

Decreasing on the interval : $\left (-\infty \right, \infty )$.

The graph of the function $f(x)=2^{-x}$ is :

Domain of the function is all real numbers.

Range of the function is $(0, \infty )$.

The function does not have any $x$ -intercept.

$y$ - intercepts is $0$

Horizontal asymptote of the function is $y=0$.

End behavior : $\lim_{x\rightarrow -\infty }(f(x))=\infty$ and $\lim_{x\rightarrow \infty }f(x)=0$.

The function is decreasing over the interval : $\left (-\infty \right, \infty )$.

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