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3 years ago

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State the domain restriction(s) in interval notation of \displaystyle f\left(g\left(x\right)\right)f(g(x)) given: \displaystyle

f\left(x\right)=\sqrt{3x-2}f(x)= 3x−2 and \displaystyle g\left(x\right)=x-7g(x)=x−7

1 answer:

DENIUS [597]3 years ago

7 0

Answer:

The interval notation for the domain is $[\frac{23}{3},\infty ]$ .

Step-by-step explanation:

Consider the provided information.

It is given that $\:f\left(x\right)=\sqrt{3x-2},\:\text{ and }\:g\left(x\right)=x-7$

We need to find the value of $f\left(g\left(x\right)\right)$ .

Put the value of g(x) in $f\left(g\left(x\right)\right)$ .

$f\left(g\left(x\right)\right)=f(x-7)$ ....(1)

Now, put x=x-7 in $\:f\left(x\right)=\sqrt{3x-2}$

$\:f\left(x-7\right)= \sqrt{3(x-7)-2}$

$\:f\left(x-7\right)= \sqrt{3x-21-2}$

$\:f\left(x-7\right)= \sqrt{3x-23}$

From equation 1.

$f\left(g\left(x\right)\right)=\:f\left(x-7\right)= \sqrt{3x-23}$

The domain of the function is the set of input values for which a function is defined.

Here, the value of $3x-23$ should be greater or equal to 0 as the square root of a negative number is not real.

Domain= $3x-23\geq0$

$x\geq\frac{23}{3}$

The value of x is all real number greater than $\frac{23}{3}$ .

Hence, the interval notation for the domain is $[\frac{23}{3},\infty ]$ .

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Step-by-step explanation:

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Step-by-step explanation:

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Part 2

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$\chi^2 =\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

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