Choose the correct domain of the function in interval notation.

Mathematics

2 answers:

neonofarm [45]3 years ago

8 0

Answer:

domain : (-∞,∞)

Step-by-step explanation:

Domain is the set of x values for which the function is defined

To find domain we check the x value for which the function is undefined

In the give graph , the graph is continuous and smooth.

There is no break in the graph.

We have graph for all values of x

So domain is set of all real values

domain : (-∞,∞)

kari74 [83]3 years ago

3 0

The domain of a function is where the x of the function starts and finishes , but as that function doenst have an end or beginning, its domain is a) from negative infinity to infinity

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olchik [2.2K]

Answer:

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Step-by-step explanation:

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3 years ago

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HELP PLEASE!! I DONT UNDERSTAND!!!!!!!!!! THANKS SO MUCH

8090 [49]

Hello, please consider the following.

We will multiply the numerator and denominator by

$4+\sqrt{6x}$

to get rid of the root in the denominator.

First of all, we cannot divide by 0, right? So, we need to make sure that the denominator is different from 0.

$4-\sqrt{6x} =0\sqrt{6x}=4\\\\\text{Take the square}\\\\6x=4^2=16\\\\x=\dfrac{16}{6}=\dfrac{8}{3}$

We need to take any x real number different from 8/3 then and simplify the expression.

Let's do it!

$\begin{aligned}\dfrac{4}{4-\sqrt{6x}}&=\dfrac{4(4+\sqrt{6x})}{(4+\sqrt{6x})(4-\sqrt{6x})}\\\\&=\dfrac{4(4+\sqrt{6x})}{(4^2-\sqrt{6x}^2)}\\\\&=\dfrac{4(4+\sqrt{6x})}{(16-6x)}\\\\&=\dfrac{2(4+\sqrt{6x})}{(8-3x)}\\\\&\large \boxed{=\dfrac{8+2\sqrt{6x}}{8-3x}}\end{aligned}$