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Hunter-Best [27]
4 years ago
9

Let f(x) = x + 3 and g(x) = 1/x The graph of (gºf)(x) is shown below what is the range of ( gof) (x) .

Mathematics
1 answer:
charle [14.2K]4 years ago
4 0

Answer:

The range of the function (g\°f)(x) is

B.  is all real numbers except y=0

Step-by-step explanation:

Given functions:

f(x)=x+3

g(x)=\frac{1}{x}

To find the range of (g\°f)(x).

Solution:

In order to find (g\°f)(x) , we will plugin f(x) in function g(x).

(g\°f)(x)=g(f(x))

(g\°f)(x)=\frac{1}{x+3}

The graph of the function (g\°f)(x) shows that

1) As x approaches -3 (but never touches the line x=-3), y tends to positive or negative infinity.

2) As y approaches 0 (but never touches the line y=0) , x tends to positive or negative infinity.

Thus, the range of the function is all real numbers except y=0

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Answer:

a) \sqrt{61 - 24 \sqrt{5} }  =  - 4  + 3 \sqrt{5}

b)( \sqrt{ ( {c}^{2}   -  1) ({b}^{2}    -  1) } - {2 \sqrt{bc} }) (\sqrt{ ( {c}^{2}   -  1) ({b}^{2}    -  1) }  + {2 \sqrt{bc}  } )

c) \frac{ \sqrt{9 - 4 \sqrt{5} } }{2 -  \sqrt{5} }  =   - 1

Step-by-step explanation:

We want to simplify

\sqrt{61 - 24 \sqrt{5} }

Let :

\sqrt{61 - 24 \sqrt{5} }  = a - b \sqrt{5}

Square both sides of the equation:

(\sqrt{61 - 24 \sqrt{5} } )^{2}  =  ({a - b \sqrt{5} })^{2}

Expand the RHS;

61 - 24 \sqrt{5} =  {a}^{2}  - 2ab \sqrt{5}  + 5 {b}^{2}

Compare coefficients on both sides:

{a}^{2}  + 5 {b}^{2}  = 61 -  -  - (1)

- 24 =  - 2ab \\ ab = 12 \\ b =  \frac{12}{b}  -  -  -( 2)

Solve the equations simultaneously,

\frac{144}{ {b}^{2} }  + 5 {b}^{2}  = 61

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Solve the quadratic equation in b²

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This implies that:

b =  \pm3 \: or \: b =  \pm  \frac{4 \sqrt{5} }{5}

When b=-3,

a =  - 4

Therefore

\sqrt{61 - 24 \sqrt{5} }  =  - 4  + 3 \sqrt{5}

We want to rewrite as a product:

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as a product:

We rearrange to get:

{b}^{2}  {c}^{2}   -  {b}^{2}  -  {c}^{2}  + 1- 4bc

We factor to get:

{b}^{2} ( {c}^{2}   -  1)  -  ({c}^{2}   -  1)- 4bc

Factor again to get;

( {c}^{2}   -  1) ({b}^{2}   -  1)- 4bc

We rewrite as difference of two squares:

(\sqrt{( {c}^{2}   -  1) ({b}^{2}   -  1) })^{2} - ( {2 \sqrt{bc} })^{2}

We factor the difference of square further to get;

( \sqrt{ ( {c}^{2}   -  1) ({b}^{2}    -  1) } - {2 \sqrt{bc} }) (\sqrt{ ( {c}^{2}   -  1) ({b}^{2}    -  1) }  + {2 \sqrt{bc}  } )

c) We want to compute:

\frac{ \sqrt{9 - 4 \sqrt{5} } }{2 -  \sqrt{5} }

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\sqrt{9 - 4 \sqrt{5} }  = a - b \sqrt{5}

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Compare coefficients in both equations;

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and

- 2ab =  - 4 \\ ab = 2 \\ a =  \frac{2}{b}  -  -  -  - (2)

Put equation (2) in (1) and solve;

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