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PolarNik [594]
3 years ago
6

(b)

Mathematics
1 answer:
ElenaW [278]3 years ago
7 0

Answer:

Step-by-step explanation:

a) r = √(1² + (-5²)) = √26 = 5.09901...

   θ = tan⁻¹(-5/1) = 4.9097... radians

   (5.1, 4.9)

b) r = - 5.09901...

   θ = 4.9097... - π = 1.76819...

   (-5.1, 1.8)

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A. Use composition to prove whether or not the functions are inverses of each other. B. Express the domain of the compositions u
Kryger [21]

Given: f(x) = \frac{1}{x-2}

           g(x) = \frac{2x+1}{x}

A.)Consider

f(g(x))= f(\frac{2x+1}{x} )

f(\frac{2x+1}{x} )=\frac{1}{(\frac{2x+1}{x})-2}

f(\frac{2x+1}{x} )=\frac{1}{\frac{2x+1-2x}{x}}

f(\frac{2x+1}{x} )=\frac{x}{1}

f(\frac{2x+1}{x} )=1

Also,

g(f(x))= g(\frac{1}{x-2} )

g(\frac{1}{x-2} )= \frac{2(\frac{1}{x-2}) +1 }{\frac{1}{x-2}}

g(\frac{1}{x-2} )= \frac{\frac{2+x-2}{x-2} }{\frac{1}{x-2}}

g(\frac{1}{x-2} )= \frac{x }{1}

g(\frac{1}{x-2} )= x


Since, f(g(x))=g(f(x))=x

Therefore, both functions are inverses of each other.


B.

For the Composition function f(g(x)) = f(\frac{2x+1}{x} )=x

Since, the function f(g(x)) is not defined for x=0.

Therefore, the domain is (-\infty,0)\cup(0,\infty)


For the Composition function g(f(x)) =g(\frac{1}{x-2} )=x

Since, the function g(f(x)) is not defined for x=2.

Therefore, the domain is (-\infty,2)\cup(2,\infty)



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