6.3E Exercises

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Definition True/False

Answer True/False. If False, explain why.

The function $ f(x) = x^{\frac{1}{2} }$ is exponential with base $ \frac{1}{2} $.
The function $ f(x) = \left( \frac{1}{2} \right)^x $ is exponential with base $ \frac{1}{2} $.
The expression $ \left( \frac{1}{2} \right)^x $ is equivalent to the expression $ \frac{1}{2^x} $.
The function $ f(x) = 5^x $ is exponential with base 5.
The function $ f(x) = 5^{-x} $ is exponential with base 5.
The function $ f(x) = 5^{-x} $ is exponential with base $ \frac{1}{5} $.
The function $ f(x) = 3^x$ is increasing.
The function $ f(x) = \left(\frac{1}{3}\right)^x $ is increasing.
The function $ f(x) = \left(\frac{1}{3}\right)^x $ is decreasing.
For $ f(x) = 3^x$, we have $f(0) = 3 $.
For $ f(x) = 3^x$, we have $f(0) = 1 $.
For $ f(x) = 3^x$, we have $f(1) = 3 $.
The function $ f(x) = e^x $ can take negative values. That is, there exists some $x$ such that $f(x) < 0 $.
The function $ f(x) = 3^x$ can take negative values.

Answer

F, the variable isn't in the exponent.
T
T
T
F, rewrite $ 5^{-x} $ as $ \left( \frac{1}{5} \right)^x $ to see the base.
T
T
F, when the base is less than 1, the function is decreasing (downhill behavior).
T
F, $ f(0) = 3^0 = 1 $.
T
T
F, exponential functions like this never take negative values (stay above the $x$-axis).
F, same reason.

Graphs True/False

Answer True/False. If False, justify or find the error.

The graph of $f(x) = a^x$ passes through $ (0,1)$ for any base with $a > 0$ and $a \neq 1$.
The graph of $ f(x) = 2^x$ is increasing.
The graph of $ f(x) = \left( \frac{4}{5} \right)^x $ is decreasing.
The graph of $ f(x) = \left( \frac{5}{4} \right)^x $ is decreasing.
As $x \rightarrow \infty$, the graph of $ f(x) = 5^x$ is steeper (faster growing) than the graph of $g(x) = 10^x $.
As $x \rightarrow \infty$, the graph of $ g(x) = 10^x$ is steeper (faster growing) than the graph of $ f(x) = 5^x$.
The graph of $ f(x) = e^x$ is increasing.
As $x \rightarrow \infty$, the graph of $f(x) = e^x $ is steeper than the graph of $g(x) = 3^x $, for $x > 0$.
The graph of $ f(x) = 6^x $ passes through $ (1,6) $.
The graph of $ f(x) = 6^x $ passes through $ \left(-1, \frac{1}{6} \right) $.
The graph of $f(x) = 6^x $ passes through $ (6, 1) $.
The graph of $ f(x) = e^x$ passes below the $x$-axis.
The graph of $f(x) = a^x $ passes through $ (1,a)$ for any base $a$.
The graph of $ f(x) = 5^{-x} $ is the reflection across the $y$-axis of the graph of $ g(x) = 5^x$.
The graph of $ f(x) = \left( \frac{1}{7} \right)^x $ is the reflection across the $y$-axis of the graph of $ g(x) = 7^x $.

Answer

T
T
T
F, the base $ \frac{5}{4} > 1 $ so it's increasing.
F, bigger base gives faster growth.
T
T
F, $e \approx 2.71...$ is less than 3, so $g$ has the larger base and thus faster growth.
T
T
F, $ f(6) = 6^6 $.
F, as mentioned, these exponential functions are always positive.
T
T
T

Matching Graphs

By analyzing signal points like $ (1,a)$ and $ \left(-1, \frac{1}{a}\right) $ and using your knowledge of exponential function graphs, match the graphs to their functions.

$z1.png$	$z4.png$	$z2.png$	$z3.png$
1.	2.	3.	4.
$ f(x) = e^x$	$ g(x) = \left(\frac{1}{3}\right)^x $	$ h(x) = 7^x$	$ p(x) = 4^{-x} $

Answer

$h$
$p$
$f$
$ g$

Working With Exponential Functions

(No calculator) Fill in the table of values for the function $ f(x) = 2^x $.

x	$ f(x) $	x	$ f(x)$
1		0
2		-1
3		-2

(No calculator) Fill in the table of values for the function $ f(x) = \left(\frac{2}{3}\right)^x$.

x	$ f(x) $	x	$ f(x)$
1		0
2		-1
3		-2

(No calculator) Fill in the table of values for the function $ f(x) = 5^{-x} $.

x	$ f(x) $	x	$ f(x)$
1		0
2		-1
3		-2

For $ f(x) = 5^x $, what should $x$ be to get $f(x) = 5$? What should $x$ be to get $f(x) = 25$?
For $ f(x) = 5^x$, what should $x$ be to get $ f(x) = \frac{1}{5} $? What should $x$ be to get $ f(x) = 1$?
For $f(x) = 4^x$, what should $x$ be to get $f(x) = 2$?
For $f(x) = 8^x$, what should $x$ be to get $f(x) = 2$?
If $f(x) = 2^x$ and $g(x) = 4^x$, is there an $x$-value for which $f(x) = g(x)$?
Is there an $x$-value such that $ e^x = -2 $?

Using technology (calculator or WolframAlpha, for example), fill in the table of values for the function $f(x) = (1.2)^x $.

x	$ f(x) $	x	$ f(x) $
-2		0
-1		3.25
-0.5		10

Answer

x	$ f(x) $	x	$ f(x)$
1	2	0	1
2	4	-1	$ \frac{1}{2}$
3	8	-2	$ \frac{1}{4} $

x	$ f(x) $	x	$ f(x)$
1	$ \frac{2}{3} $	0	1
2	$ \frac{4}{9} $	-1	$ \frac{ 3}{2} $
3	$ \frac{8}{27} $	-2	$ \frac{9}{4} $

x	$ f(x) $	x	$ f(x)$
1	$ \frac{1}{5} $	0	1
2	$ \frac{1}{25} $	-1	5
3	$ \frac{1}{125} $	-2	25

For $f(x) = 5$, $x$ should be 1. For $f(x) = 25$, $x $ should be 2.
For $ f(x) = \frac{1}{5} $, $x$ should be $-1$. For $ f(x) = 1$, $x$ should be 0.
We want $x$ such that $ f(x) = 4^x = 2 $. What kind of power should 4 be raised to to get out...its square root? That's right, $x = \frac{1}{2} $. Check: $ f(1/2) = 4^{\frac{1}{2}} = \sqrt{4} = 2 $.
Similar reasoning, $x = \frac{1}{3} $.
Yes, both pass through $ (0,1)$. Aka, if $x = 0$, then $f(x) = g(x) = 1$.
No, there is no possible input that will make $f(x) = e^x$ pass below the $x$-axis into negative function values.

Let me teach you a technology trick for this kind of thing (for when you're allowed to use technology, of course, not for cheating). Go to wolframalpha.com and type in "table of values (1.2)^x for x = -2, -1, -0.5, 0, 3.25, 10" and hit Enter. You're welcome.

x	$ f(x) $	x	$ f(x) $
-2	0.694444	0	1
-1	0.833333	3.25	1.80859
-0.5	0.912871	10	6.19174

Applications

1. Unless you receive a raise every year equivalent to the rate of inflation, you are essentially taking a wage cut as the purchasing power of your money goes down! My dad's current salary is $100,000 per year. If inflation is currently steady at a rate of 4% per year, and his employer promises to give him a fair cost-of-living raise to match that each year, what does he expect his annual salary to be after 2 years? After 10 years? Write a function that gives his salary, $S$, as a function of how many years, $y$, have passed. You can use a calculator to perform the computations.

2. My video is going marginally viral on social media right now. Right before it took off, it was sitting at only 40 views, but now the number of views is doubling every minute! Assume the algorithm blesses me for ten minutes with this growth. After 2 minutes, how many views does the video have? After 5 minutes? After the 10 minutes? Write a function giving the total number of views on my video as a function of how many minutes have passed.

Answer

1. After one year, his salary will be raised to $ 100000(1.04) = 104000 $. After two years, it would be $ 100000(1.04)^2 = 108160$. Following the pattern, after ten years it will be $ 100000(1.04)^{10} = 148024 $. The function is $ S(y) = 100000(1.04)^y$.

2. After one minute, I have $ 40(2) = 80$ views. Every time a minute passes, I must multiply by 2 to effectively double the count. The function is thus $ N(t) = 40\cdot 2^t $. At $t = 2$, $N = 160$ views. At $t = 5$, $N = 1280$. At $t = 10$, $N \approx 41$k views.

x	\( f(x) \)	x	\( f(x)\)
1	2	0	1
2	4	-1	\( \frac{1}{2}\)
3	8	-2	\( \frac{1}{4} \)

x	\( f(x) \)	x	\( f(x)\)
1	\( \frac{2}{3} \)	0	1
2	\( \frac{4}{9} \)	-1	\( \frac{ 3}{2} \)
3	\( \frac{8}{27} \)	-2	\( \frac{9}{4} \)

x	\( f(x) \)	x	\( f(x)\)
1	\( \frac{1}{5} \)	0	1
2	\( \frac{1}{25} \)	-1	5
3	\( \frac{1}{125} \)	-2	25

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