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2 years ago

The graph of the function f(x)=V x+4 is shown below The domain of the function is

Mathematics

1 answer:

ladessa [460]2 years ago

3 0

Answer:

The function domain is:

{x|x ≥ -4}

Hence, option 4 is correct.

Step-by-step explanation:

Given the expression of the function

$f\left(x\right)=\sqrt{x+4}$

We know that the domain of the function is the set of input values for which the function is real and defined

We know that the value inside the radical expression must be greater than or equal to 0 so that the function remains defined.

$x+4\ge \:0$

subtract 4 from both sides

$x+4-4\ge \:0-4$

simplify

$x\ge \:-4$

Therefore, the function domain is:

{x|x ≥ -4}

Hence, option 4 is correct.

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valentina_108 [34]

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Step-by-step explanation:

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2 years ago

<img src="https://tex.z-dn.net/?f=%20%7B2%7D%5E%7Bx%7D%20%20-%20%20%20%7B3%7D%5E%7Bx%7D%20%20%3D%20%20%20%20%5Csqrt%7B%20%7B6%7D

strojnjashka [21]

Answer:

x = 0

Step-by-step explanation:

${2}^{x} - {3}^{x} = \sqrt{ {6}^{x} } - \sqrt{ {9}^{x } } \\ \\ {2}^{x} - {3}^{x} = \sqrt{ {6}^{x} } - \sqrt{ {( {3}^{2}) }^{x } } \\ \\ {2}^{x} - {3}^{x} = \sqrt{ {6}^{x} } - \sqrt{ {({3)}^{2x}}}\\ \\ {2}^{x} - \cancel{{3}^{x}} = \sqrt{ {6}^{x} } - \cancel{ {3}^{x} } \\ \\ {2}^{x} = \sqrt{ {6}^{x} } \\ squaring \: both \: sides \\ \\ {( {2}^{x} )}^{2} = {(\sqrt{ {6}^{x} })}^{2} \\ \\ {2}^{2x} = {6}^{x} \\ \\ {2}^{2x} = {(2 \times 3)}^{x} \\ \\ {2}^{2x} = {2 ^{x} \times 3}^{x} \\ \\ \frac{ {2}^{2x} }{ {2}^{x} } = 3^{x} \\ \\ {2}^{2x - x} = {3}^{x} \\ \\ {2}^{x} = {3}^{x} \\ \\ \frac{ {2}^{x} }{ {3}^{x} } = 1 \\ \\ \bigg( \frac{2}{3} \bigg)^{x} = 1 \\ \\ \implies \: x = 0 \\ \\ \because \: for \: x = 0 \\ \\ \bigg( \frac{2}{3} \bigg)^{0} = 1$

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2 years ago