find the inverse of f(x)=(1/2x+2)sq Determine whether it is function and state its domain and range.

Question

step  by  step

Answer 1

Given function f(x) = (1/2x+2)

y = √(1/2x+2)

Squringon each side.

y^2 = 1/2x+2

Subtract 2 from each side.

y^2-2 = 1/2x+2-2

y^2-2 = 1/2x

Multiple to each side by 2.

2(y^2-2) = 2*1/2x

2y^2-4 = x

Inverse of the function is x = 2y^2-4

 

 

Answer 2

• The function is .

Interchange x and y.

Squaring on each side.

.

Therefore, inverse of the function is .

• The function .

The domain of a function is for all values of x greater than or equals to - 4, makes the function mathematically correct.

Since there shouldn't be any negative numbers in the square root.

So, the domain of the above function is { x | x ≥ - 4}

DOMAIN : .

Range set is the corresponding values of the function for different values of x.

Since for for all values of x greater than or equals to - 4, the function is

So, the range of the function is always greater than or equal to .

RANGE : (all non negative real numbers).

Answer 3

The function is f(x) = √(x/2 + 2).

Find the f ^{- 1}(x) as follows :

Replace f(x) with y in the original equation.

f(x) = √(x/2 + 2) → y = √(x/2 + 2).

Interchange x and y and solve for y.

x = √(y/2 + 2)

x² = y/2 + 2

x² - 2 = y/2

2(x² - 2) = y

Replace y with  f ^{- 1}(x).

y = 2(x² - 2) → f ^{- 1}(x) = 2(x² - 2).

The inverse of f(x) = √(x/2 + 2) is f ^{- 1}(x) = 2(x² - 2) or y = 2(x² - 2).

To each value of x there corresponds exactly one value of y. So, f ^{- 1}(x) is a function of x.

Domain :

The function f ^{- 1}(x) = y = 2(x² - 2) is a quadratic function or parabola function.There are no rational or radical expressions, so there is nothing that will restrict the domain. Any real number can be used for x to get a meaningful output.

The domain of f ^{- 1}(x) = 2(x² - 2) is all real numbers.

Range :

Find the range of f ^{- 1}(x) = y = 2(x² - 2) algebraically as follows.

Compare the equation y = 2x² - 4 with quadratic or parabola function y = f(x) = ax² +bx + c.

a = 2, b = 0 and c = - 4.

a = 2 is positive number the parabola opens up and f(x) has minimum value at x = - b/2a = - 0/2(2) = 0.

The minimum value is f(- b/2a) = f(0) = [ 2(0)² - 4 ] = - 4.

The minimum value of y = f(x) = - 4,  so the graph of parabola cannot be lower than - 4.

Thus, the range of function y = f ^{- 1}(x) ≥ - 4.

Range of the function is { f ^{- 1}(x) ∈ R : f ^{- 1}(x) ≥ - 4 }.