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Andre45 [30]
3 years ago
8

CAN PLEASE PLEASE GET HELP PLEASE I NEED IT PLEASE AND THANK YOU

Mathematics
1 answer:
Katena32 [7]3 years ago
6 0

Answer:

Step-by-step explanation:

the domain for (2x-1)/(x^2-7) is x not equal to +/- sqrt(7)

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Which statement describes the inverse of m(x) = x^2 – 17x?
DochEvi [55]

Given:

The function is

m(x)=x^2-17x

To find:

The inverse of the given function.

Solution:

We have,

m(x)=x^2-17x

Substitute m(x)=y.

y=x^2-17x

Interchange x and y.

x=y^2-17y

Add square of half of coefficient of y , i.e., \left(\dfrac{-17}{2}\right)^2 on both sides,

x+\left(\dfrac{-17}{2}\right)^2=y^2-17y+\left(\dfrac{-17}{2}\right)^2

x+\left(\dfrac{17}{2}\right)^2=y^2-17y+\left(\dfrac{17}{2}\right)^2

x+\left(\dfrac{17}{2}\right)^2=\left(y-\dfrac{17}{2}\right)^2        [\because (a-b)^2=a^2-2ab+b^2]

Taking square root on both sides.

\sqrt{x+\left(\dfrac{17}{2}\right)^2}=y-\dfrac{17}{2}

Add \dfrac{17}{2} on both sides.

\sqrt{x+\left(\dfrac{17}{2}\right)^2}+\dfrac{17}{2}=y

Substitute y=m^{-1}(x).

m^{-1}(x)=\sqrt{x+(\dfrac{189}{4}})+\dfrac{17}{2}

We know that, negative term inside the root is not real number. So,

x+\left(\dfrac{17}{2}\right)^2\geq 0

x\geq -\left(\dfrac{17}{2}\right)^2

Therefore, the restricted domain is x\geq -\left(\dfrac{17}{2}\right)^2 and the inverse function is m^{-1}(x)=\sqrt{x+(\dfrac{189}{4}})+\dfrac{17}{2}.

Hence, option D is correct.

Note: In all the options square of \dfrac{17}{2} is missing in restricted domain.

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D)4 is the answer jcidifodofif
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