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Gelneren [198K]
2 years ago
8

HELP MEE!

Mathematics
1 answer:
luda_lava [24]2 years ago
5 0
The answer is B. y=5x
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If m (x) = StartFraction x + 5 Over x minus 1 EndFraction and n(x) = x – 3, which function has the same domain as (m circle n) (
Andrew [12]

Answer:

The function h (x) = StartFraction 11 Over x minus 4 EndFraction

has the domain of h (x) = ( - ∞, 4) U (4, ∞ ) which is same as the

domain of (m ∘ n)(x) = ( - ∞, 4) U (4, ∞ ).

Step-by-step explanation:

m (x) = StartFraction x + 5 Over x minus 1 and n(x) = x – 3

As m (x) could be written as:

m(x) = \frac{x+5}{x-1}

n(x) = x – 3

(m ∘ n)(x) = m(n(x)) = m(x – 3)

              = \frac{(x-3) + 5}{(x-3) - 1}

              = \frac{x+2}{x-4}

In order to find the domain of (m ∘ n)(x) = \frac{x+2}{x-4}, we need to

make sure that denominator can not be zero,

So,

     x - 4 = 0

     x = 4    

So, our domain can be anything except for 4.

Hence, Domain = D = ( - ∞, 4) U (4, ∞ )

Now, compare the domain of (m ∘ n)(x) i.e D = ( - ∞, 4) U (4, ∞ ) with all the options:

Option A: h (x) = x + 5 / 11

Option A has the Domain of h (x) = Set of all real numbers

So, Option A is false.

Option B: h (x) = 11 / x - 1

                        x - 1 = 0

                         x = 1

Option B has the domain of h (x) = ( - ∞, 1) U (1, ∞ )

So, Option B is false.

Option C: h (x) = 11 / x - 4

                        x - 4 = 0

                         x = 4

Domain C has the domain of h (x) = ( - ∞, 4) U (4, ∞ )

So, Option C is true.

Option D: h (x) = 11 / x - 3

                        x - 3 = 0

                         x = 3

Option D has the domain of h (x) = ( - ∞, 3) U (3, ∞ )

So, option D is also false.

Hence, only option C is true as the option C i.e. h (x) = StartFraction 11 Over x minus 4 EndFraction has the domain of h (x) = ( - ∞, 4) U (4, ∞ ) which is same as the domain of (m ∘ n)(x) = ( - ∞, 4) U (4, ∞ ).

keywords: Domain, composite function

Learn more about the domain of function from brainly.com/question/13020740

#learnwithBrainly

4 0
4 years ago
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