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MatroZZZ [7]
2 years ago
11

Find the domain of the function

Mathematics
1 answer:
Svetradugi [14.3K]2 years ago
7 0

Answer:

<em>The domain of f is (-∞,4)</em>

Step-by-step explanation:

<u>Domain of a Function</u>

The domain of a function f is the set of all the values that the input variable can take so the function exists.

We are given the function

f(x)=\frac{1}{\sqrt{4-x} }

It's a rational function which denominator cannot be 0. In the denominator, there is a square root whose radicand cannot be negative, that is, 4-x must be positive or zero, but the previous restriction takes out 0 from the domain, thus:

4 - x > 0

Subtracting 4:

- x > -4

Multiplying by -1 and swapping the inequality sign:

x < 4

Thus the domain of f is (-∞,4)

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\rightarrow z^4=-625\\\\\rightarrow z=(-625+0i)^{\frac{1}{4}}\\\\\rightarrow x+iy=(-625+0i)^{\frac{1}{4}}\\\\ x=r \cos A\\\\y=r \sin A\\\\r \cos A=-625\\\\ r \sin A=0\\\\x^2+y^2=625^{2}\\\\r^2=625^{2}\\\\|r|=625\\\\ \tan A=\frac{0}{-625}\\\\ \tan A=0\\\\ A=\pi\\\\\rightarrow z= [625(\cos (2k \pi+pi) +i \sin (2k\pi+ \pi)]^{\frac{1}{4}}\\\\k=0,1,2,3,4,....\\\\\rightarrow z=(625)^{\frac{1}{4}}[\cos \frac{(2k \pi+pi)}{4} +i \sin \frac{(2k\pi+ \pi)}{4}]

\rightarrow z_{0}=(625)^{\frac{1}{4}}[\cos \frac{pi}{4} +i \sin \frac{\pi)}{4}]\\\\\rightarrow z_{1}=(625)^{\frac{1}{4}}[\cos \frac{3\pi}{4} +i \sin \frac{3\pi}{4}]\\\\ \rightarrow z_{2}=(625)^{\frac{1}{4}}[\cos \frac{5\pi}{4} +i \sin \frac{5\pi}{4}]\\\\ \rightarrow z_{3}=(625)^{\frac{1}{4}}[\cos \frac{7\pi}{4} +i \sin \frac{7\pi}{4}]

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We have to find those roots among four roots whose argument is between 270° and 360°.So, that root is

   \rightarrow z_{2}=(625)^{\frac{1}{4}}[\cos \frac{5\pi}{4} +i \sin \frac{5\pi}{4}]

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