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F(x) = (sqrt x+1)-2 find the inverse

User Petro
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Answer:


f^(-1)(x)=(x+2)^2-1, \quad \{x : x \geq -2\}

Explanation:

Given function:


f(x)=√(x+1)-2

Domain: As the square root of a negative number cannot be taken, the domain of the given function is restricted to {x : x ≥ -1}.

Range: As the domain is restricted, the range of the given function is also restricted: {f(x) : f(x) ≥ -2}.

The inverse of a function its reflection in the line y = x.

To find the inverse of a function, swap f(x) for y:


\implies y=√(x+1)-2

Rearrange the equation to make x the subject:


\begin{aligned}\implies y&=√(x+1)-2\\y+2&=√(x+1)-2+2\\y+2&=√(x+1)\\(y+2)^2&=(√(x+1))^2\\(y+2)^2&=x+1\\(y+2)^2-1&=x+1-1\\x&=(y+2)^2-1\end{aligned}

Swap the x for f⁻¹(x) and the y for x:


\implies f^(-1)(x)=(x+2)^2-1

The domain of the inverse function is the range of the function.

Therefore, the domain of f⁻¹(x) is restricted to {x : x ≥ -2}.

F(x) = (sqrt x+1)-2 find the inverse-example-1
User Munin
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