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3 years ago

Which function is f(x) equal to f^-1(x)?

Mathematics

2 answers:

Morgarella [4.7K]3 years ago

6 0

Answer:

c. f(x) x+1/x-1

Step-by-step explanation:

To answer this question, we need to check each answer one by one until we find the right one.

y = (x+6)/(x-6)

switch x and y

x = (y+6)/(y-6)

solve for y

x(y-6) = y+6

xy - 6x = y+6

y(x-1) = 6x+6

y = (6x+6) /(x-1) = 6(x+1)/(x-1)

f^-1(x) = 6(x+1)/(x-1)

y = (x+2)/(x-2)

switch x and y

x = (y+2)/(y-2)

solve for y

x(y-2) = y+2

xy -2x = y+2

y(x-1) = 2x+2

y = (2x+2)/(x-1)

f^-1(x) = 2(x+1)/(x-1)

y = (x+1)/(x-1) ------ correct one

switch x and y

x = (y+1)/(y-1)

solve for y

x(y-1) = y+1

xy - x = y+1

y(x-1) = x+1

y = (x+1)/(x-1)

f^-1(x) = (x+1)/(x-1)

f(x) = f^-1(x)

hammer [34]3 years ago

6 0

Answer:

C. f(x) = x+1/x-1

Step-by-step explanation:

y = (x+1)/(x-1)

Interchange the x and the y

x = (y+1)/(y-1)

Now, we will solve for y

x(y-1) = y+1

xy - x = y+1

y(x-1) = x+1

y = (x+1)/(x-1)

f^-1(x) = (x+1)/(x-1)

f(x) = f^-1(x)

I hope I helped!

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~ Zoe

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Brut [27]

Answer:

cos 37=x/b (second one)

sin 37= a/b (third one)

tan 37=a/x (sixth one)

Step-by-step explanation:

Cos is defined by the adjacent side over the hypotenuse. Cos 37 would be x/b. Clearly, that is option 2. Sin is defined by opposite over hypotenuse. Sin 37 would be a/b. That is the 3rd option. Tan is defined by opposite over adjacent. Tan 37 would be a/x or the 6th option.

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3 years ago

Fiona is trying to prove that a pair of expressions are equivalent by using substitution. Which pair of expressions has the same

fiasKO [112]

Missing Part of the Question

3 (4 - y) -7 ; 5 + 3y

9 + y + 4y - 2 - 3y; -2y + 7

5 + y - 2 - 6y + y; 8y + 3

2y + 7 + y - 2 + y; 4y + 5

Answer:

2y + 7 + y - 2 + y; 4y + 5

Step-by-step explanation:

Given

y = 3

Required

Which pair of expression are equal

To get the pair of expression that are equal; we simply substitute 3 for y in each of the expressions

3 (4 - y) -7 and 5 + 3y

Substitute 3 for y in 3 (4 - y) -7

This gives 3(4 - 3) -7

= 3(1) - 7

= 3 - 7

= -4

Then Substitute 3 for y in 5 + 3y

This gives 5 + 3(3)

= 5 + 9

= 14

-4 and 14 are not equal

---------------------------------------------------------------------------------------

9 + y + 4y - 2 - 3y; -2y + 7

Substitute 3 for y in 9 + y + 4y - 2 - 3y

This gives 9 + 3 + 4(3) - 2 - 3(3)

= 9 + 3 + 12 - 2 - 9

= 13

Then Substitute 3 for y in -2y + 7

This gives -2(3) + 7

= -6 + 7

= 1

13 and 1 are not equal

---------------------------------------------------------------------------------------

5 + y - 2 - 6y + y; 8y + 3

Substitute 3 for y in 5 + y - 2 - 6y + y

This gives 5 + 3 - 2 - 6(3) + 3

= 5 + 3 - 2 - 18 + 3

= -9

Then Substitute 3 for y in 8y + 3

This gives 8(3) + 3

= 24 + 3

= 27

-9 and 27 are not equal

---------------------------------------------------------------------------------------

2y + 7 + y - 2 + y; 4y + 5

Substitute 3 for y in 2y + 7 + y - 2 + y

This gives 2(3) + 7 + 3 - 2 + 3

= 6 + 7 + 3 - 2 + 3

= 17

Then Substitute 3 for y in 4y + 5

This gives 4(3) + 5

= 12 + 5

= 17

17 and 17 are equal

Hence, Option D is correct

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3 years ago

If m(-12) =0 is the result of M(x) using the reminder theorem then what do -12 and 0 represent?

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Suppose we wish to determine whether or not two given polynomials with complex coefficients have a common root. Given two first-degree polynomials a0 + a1x and b0 + b1x, we seek a single value of x such that

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Hence a non-trivial solution requires the vanishing of the determinant of the coefficient matrix, which again gives a0b1 - a1b0 = 0.

Now consider two polynomials of degree 2. In this case we seek a single value of x such that

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g A scientist testing the effects of a chemical on apple yield sprays an orchard with the chemical. A second orchard does not re

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Step-by-step explanation:

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3 years ago

Write one sine and one cosine equation for each graph below.

pychu [463]

Answer:

Q13. y = sin(2x – π/2); y = - 2cos2x

Q14. y = 2sin2x -1; y = -2cos(2x – π/2) -1

Step-by-step explanation:

Question 13

(A) Sine function

y = a sin[b(x - h)] + k

y = a sin(bx - bh) + k; bh = phase shift

(1) Amp = 1; a = 1

(2) The graph is symmetrical about the x-axis. k = 0.

(3) Per = π. b = 2

(4) Phase shift = π/2.

2h =π/2

h = π/4

The equation is

y = sin[2(x – π/4)} or

y = sin(2x – π/2)

B. Cosine function

y = a cos[b(x - h)] + k

y = a cos(bx - bh) + k; bh = phase shift

(1) Amp = 1; a = 1

(2) The graph is symmetrical about the x-axis. k = 0.

(3) Per = π. b = 2

(4) Reflected across x-axis, y ⟶ -y

The equation is y = - 2cos2x

Question 14

(A) Sine function

(1) Amp = 2; a = 2

(2) Shifted down 1; k = -1

(3) Per = π; b = 2

(4) Phase shift = 0; h = 0

The equation is y = 2sin2x -1

(B) Cosine function

a = 2, b = -1; b = 2

Phase shift = π/2; h = π/4

The equation is

y = -2cos[2(x – π/4)] – 1 or

y = -2cos(2x – π/2) - 1

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3 years ago