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vivado [14]
2 years ago
9

How to solve these 3 problems

Mathematics
1 answer:
Bingel [31]2 years ago
3 0

Answer:

  1. a. decay; b. growth; c. decay; d. neither
  2. r = 4; a = 1; y = 1·4^x
  3. a. an = 3(5^(n-1)); b. f(x) = (3/5)(5^x); c. exponential growth; d. y-intercept: 3/5; first term: 3.

Step-by-step explanation:

There are two kinds of exponential problems here.

  1. exponential functions of the form f(x) = a·b^x
  2. exponential sequences of the explicit form an = a1·r^(n-1)

The second problem gives you a table that suggests the sequence form, but it asks for the exponential function form. The third problem does something similar.

__

<h3>1.</h3>

In an exponential function of the form f(x) = a·b^x, the function grows if b>1 and decays if b<1. Using this check, we can easily answer ...

  a. 0.4 < 1 . . . decay

  b. 1.3 > 1 . . . growth

  c. 1/2 < 1 . . . decay

  d. 1 = 1 . . . neither growth nor decay; the function is constant: j(x) = 1.

__

<h3>2.</h3>

The value of x is given starting at 1, so we can consider this a geometric sequence. The common ratio is r = 16/4 = 4. The first term is a1 = 4, so the explicit formula for the sequence is ...

  an = 4·4^(n-1)

When this is expanded to get rid of the constant in the exponent, we have ...

  an = 4·(4^n)·(4^-1) = 1·4^n

We recognize this form as matching the functional form f(x) = a·r^x. The multiplier of the exponential factor is a=1. In summary, ...

  r = 4; a = 1; f(x) = 1·4^x

__

<h3>3.</h3>

The first term of this geometric sequence is a1 = 3. The common ratio is r = 15/3 = 5. Using the explicit formula, we have ...

 a. explicit form: an = 3·5^(n-1)

Using the method of question 2 to write the functional form, we find ...

  an = 3(5^n)(5^-1) = (3/5)(5^n)

  b. functional form: f(x) = (3/5)(5^x)

  c. function family: exponential growth functions

 d. y-intercept: (3/5) . . . . read this from the f(x) form

     1st term: the first term listed in the given sequence is 3

_____

<em>Additional comment</em>

The "y-intercept" of a sequence is irrelevant (undefined), as the sequence term numbering starts with 1, not 0. The domain of the explicit formula is <em>natural numbers</em>, which does not include 0.

Similarly, the "first term" of a function f(x) needs further definition. Here, we've answered the question by saying the first term is f(1). There is no conventional definition of a "first term" for a continuous function.

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Given the function f(x) = 2|x + 6| – 4, for what values of x is f(x) = 6?
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Step-by-step explanation:

| x | = x

|- x | = x

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x = 1 , f( 1 ) = 2 | 1 + 6 | - 4 = 2 | 7 | - 4 = 14 - 4 = 10 , False

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x = 14 , f( 14 ) = 2 | 14 + 6 | - 4 = 2 | 20 | - 4 = 40 - 4 = 36 , False

x = -14 , f( - 14 ) = 2 | - 14 + 6 | - 4 = 2 | - 8 | - 4 = 16 - 4 = 12 , False

x = - 26 , f( - 26) = 2 | -26 + 6 | - 4 = 2 | -20 | - 4 = 40 - 4 = 36 , False

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                                                    OR

Given f ( x ) = 6

f( x ) = 2 | x + 6 | - 4

6  = 2 | x + 6 | - 4

6 + 4 = 2 | x + 6| - 4 + 4                 [ adding 4 to both sides ]

10 = 2 | x + 6 | + 0

5 = | x + 6 |                                    [ dividing both sides by 2 ]

- 5 =\  ( x+ 6) \ = 5\\\\                            [ \ | x | = a\  => \ -a \ =  x \ = a  \ ]

- 5 - 6 =  \ x + 6 - 6 \ = \ 5 - 6\\            [ \ subtracting \ by \ 6 \ ]

-11 = \ x \ =  - 1

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