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Restrict the domain of the quadratic function and find its inverse. Confirm the inverse relationship using composition. f(x) = 0.2x2 The domain is (x| x ). Since f−1f(x) = x for x 0, it has been confirmed that f−1(x) = for x ≥ 0 is the inverse function of f(x) = 0.2x2 for x 0. Chose the graph of the function and its inverse.

Respuesta :

The given function is

f(x)= 0.2 x²

Since f(x) will be defined for all real values of x.

So, Domain of f(x) will be ( x| x is a real number.)→This is set builder notation.

Finding the inverse of f(x):

y = 0.2 x²

→ x²= 5 y

→x = [tex]\pm\sqrt{5 y}[/tex]→ → Inverse of f(x)

Replacing x by y and y by x,we get inverse of the given function

y = [tex]\pm\sqrt{5 x}[/tex]→ →Domain x ≥ 0, x∈[0,∞]

Graph of function and its inverse are shown below.

Ver imagen Аноним

Answer:

Domain = {x | x≥0 }

Inverse = f⁻¹(x) = [tex]\sqrt 5y[/tex].

Step-by-step explanation:

The given function is [tex]f(x)=0.2x^{2}[/tex]

We know that the domain of all functions is the whole real line.

But as [tex]x^{2}\geq 0[/tex].

So, the domain of f(x) is the positive real line.

Thus, the restriction to the domain of f(x) is {x | x≥0 }.

Now, we will find the inverse of f(x),

[tex]y=0.2x^{2}[/tex]

i.e. [tex]x^{2}=\frac{y}{0.2}[/tex]

i.e. [tex]x^{2}=5y[/tex]

i.e. [tex]x=\sqrt 5y[/tex]

Hence, the inverse of f(x) is f⁻¹(x) = [tex]\sqrt 5y[/tex].

Further, we will check the inverse using composition rule.

i.e. fοf⁻¹(x) = f⁻¹οf(x)

i.e. f(f⁻¹(x)) = f⁻¹(f(x))

i.e. f([tex]\sqrt 5y[/tex]) = f⁻¹([tex]0.2x^{2}[/tex])

i.e. [tex]0.2(\sqrt{5x})^{2}[/tex] = [tex]\sqrt{5\times 0.2x^{2}}[/tex]

i.e. 0.2 × 5x = [tex]\sqrt{x^{2}}[/tex]

i.e. x = x

Hence, we get that the function [tex]f(x)=0.2x^{2}[/tex] has inverse f⁻¹(x) = [tex]\sqrt 5y[/tex].

The graph of the function and its inverse can be seen below.

Ver imagen wagonbelleville

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