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3 years ago

If f(x) = x^2 -1, what is the equation for f–1(x)?

Mathematics

2 answers:

schepotkina [342]3 years ago

5 0

Answer:

The inverse is ±sqrt(x+1)

Step-by-step explanation:

y = x^2 -1

Exchange x and y

x = y^2 -1

Solve for y

Add 1 for each side

x+1 = y^2-1+1

x+1 = y^2

Take the square root of each side

±sqrt(x+1) = sqrt(y^2)

±sqrt(x+1) = y

The inverse is ±sqrt(x+1)

sweet [91]3 years ago

5 0

Answer:

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

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

Step-by-step explanation:

y = x² - 1

x² = y + 1

x = +/- sqrt(y + 1)

Interswitch variables:

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

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

Where sqrt: square root

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Answer:

$\frac{7}{8}$

Step-by-step explanation:

In any circle the following ratios are equal

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2 years ago

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3 years ago

M - 1 = 2 what do I need to do to solve.

andreyandreev [35.5K]

In order to solve this, start by adding 1 to both sides of the equation.

m-1=2

+1 +1

-----------

m=3

The solution to your equation is 3.

You can also check to see if this is true by replacing m with 3.

m-1=2

3-1=2

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2 years ago

Misha Larkins [42]

12 beacuse he has enough to to it to ur soo good for the game to play it and ur soo fun

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2 years ago

If a < 0 and c > b, circle the expression that is greater:

matrenka [14]

Answer:

The expression that is greater is a (b - c)

Step-by-step explanation:

a < 0 and c > b, this means (by the addition property) that c - c > b - c⇒0 > b - c

so for the product a(b - c) we would have a multiplication of a negative number a and another negative number (b - c). We know that the result of the multiplication of two negative numbers is a positive number.

Therefore, a (b - c) > 0

a < 0 and c > b, this means by the addition property that c - b > b - b⇒ c - b > 0

so for a(c - b), we have the negative number a multiplied by the positive number (c - b). We know that the result of the multiplication of a negative number by a positive number is negative.

Therefore a (c - b) < 0

Thus, the expression that is greater is the positive one which is a (b - c)

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2 years ago