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tino4ka555 [31]
3 years ago
15

Find the indicated limit, if it exists. limit of f of x as x approaches 9 where f of x equals x plus 9 when x is less than 9 and

f of x equals 27 minus x when x is greater than or equal to 9
The limit does not exist.
18
0
9

Mathematics
1 answer:
SVEN [57.7K]3 years ago
8 0

Answer:

B. 18

Step-by-step explanation:

For the function

f(x)=\left\{\begin{array}{l}x+9,\ \ x

we can find the value of the function for all x that are very close to 9 but are less than 9 and for all values of x that are very close to 9 but are greater than 9.

1. For x

\lim \limits_{x\rightarrow 9}f(x)=\lim \limits_{x\rightarrow 9}(x+9)=9+9=18

2. For x\ge 9:

\lim \limits_{x\rightarrow 9}f(x)=\lim \limits_{x\rightarrow 9}(27-x)=27-9=18

So, limit exists and is equal to 18.

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Both problems give you a function in the second column and the x-values. To find out the values of a through f, you need to plug in those x-values into the function and simplify! 

You need to know three exponent rules to simplify these expressions:
1) The negative exponent rule says that when a base has a negative exponent, flip the base onto the other side of the fraction to make it into a positive exponent. For example, 3^{-2} =
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2) Raising a fraction to a power is the same as separately raising the numerator and denominator to that power. For example, (\frac{3}{4}) ^{3}  =  \frac{ 3^{3} }{4^{3} }.
3) The zero exponent rule<span> says that any number raised to zero is 1. For example, 3^{0} = 1.
</span>

Back to the Problem:
Problem 1 
The x-values are in the left column. The title of the right column tells you that the function is y =  4^{-x}. The x-values are:
<span>1) x = 0
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2) x = 2
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Problem 2
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<span>1) x = 0
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<span>
2) x = 2
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<span>
-------

Answers: 
a = 1
b = </span>\frac{1}{16}<span>
c = </span>\frac{1}{256}
d = 1
e = \frac{4}{9}
f = \frac{16}{81}
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Anestetic [448]

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Step-by-step explanation:

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