f is a function on RR where f(x)=|x^2-2|, How do you find f^-1(x)?

2 Answers
Feb 5, 2018

Look below

Explanation:

first, we have a detail that the function f is present for RR.

the function is \abs{x^2-2}, so let's make a table from x >= 0 to x <= 2

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now graph the points, (0, 2), (1, 1), (2, 2)

graph{\abs{x^2-2}[-10, 10, -5, 5]}

now, the inverse of an absolute value function is basically the same function, which is abs{ x^2-2}

Feb 5, 2018

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Explanation:

Let's turn f(x)=abs(x^2-2) into a piece-wise function.

Now, we see that absa=a when a>=0
Similarly, we see that absa=-a when a<0

Therefore, we have:
f(x)=x^2-2 when x^2-2>=0

f(x)=-(x^2-2) when x^2-2<0

Let's graph the two parabolas. When we do, we get:

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Now, the blue parabola applies when x^2-2>=0, and green parabola applies when x^2-2<0.

Therefore, we have(focus on the red graph):
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Now, to find its inverse, we have to reflect the red graph over the line y=x.

Let's try this mathematically.

Our piece-wise function was:
f(x)=x^2-2 when x^2-2>=0

f(x)=-(x^2-2) when x^2-2<0

Let's find the inverse function for each situation.

Now, remember that f(f^-1(x))=x and f^-1(f(x))=x

Therefore, we can now find the inverse functions.

f(x)=x^2-2=>x=(f^-1(x))^2-2

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

=>+-sqrt (x+2)=f^-1(x) when x^2-2>=0

Similarly,

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

=>x-2=-(f^-1(x))^2

=>2-x=(f^-1(x))^2

=>+-sqrt (2-x)=f^-1(x) when x^2-2<0

We now graph these two sideways parabolas:
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When y>=sqrt2 or y<= -sqrt2 , then we apply our first function.

When -sqrt2<y< sqrt2 , then we apply our second function.

We therefore have:

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Just note that +-sqrt(x+2) and other relations like this one is not a function but a relation because there are more than one output for a given input.