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Step-by-step Answer

PAGE: 166

SET: Exercises

PROBLEM: 1

Please look in your text book for this problem Statement

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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