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# C3-Help with functions, please. watch

1. The function f(x) is defined by f(x)=x^2-3

a) Find f^-1(x) in similar terms:

Let y=f(x)

y=x^2-3
y+3=x^2
x=(y+3)^0.5

Therefore, f^-1(x)=(x+3)^0.5

b) Find values of x such that f(x)=f^-1(x)

I can't do this, any help would be appreciated.

The textbook says When f(x)=f^-1(x)

f(x)=x

I don't understand this.
2. (Original post by Deep456)

The textbook says When f(x)=f^-1(x)

f(x)=x

I don't understand this.
Imagine you're sketching the graph of y=f(x) then on the same axes, you're sketching y=f^-1(x). You should find that the two curves are reflections of each other in the line y=x. This is a property of inverse functions that you would have learned before.

Solving f(x)=f^-1(x) means finding where the two curves cross. They can only cross each other at the y=x line, hence if f(x)=f^-1(x) then f(x)=x.
3. (Original post by ttoby)
Imagine you're sketching the graph of y=f(x) then on the same axes, you're sketching y=f^-1(x). You should find that the two curves are reflections of each other in the line y=x. This is a property of inverse functions that you would have learned before.

Solving f(x)=f^-1(x) means finding where the two curves cross. They can only cross each other at the y=x line, hence if f(x)=f^-1(x) then f(x)=x.
Does that also mean f^-1(x)=x?

Also if you weren't using this way and doing:

x^2-3=(x+3)^0.5

How would you go about it?
4. (Original post by Deep456)
Does that also mean f^-1(x)=x?
Yes. But in this case it's easier to use the result f(x)=x because in the next line of workings, you get a quadratic to solve which is fairly straightforward.
Also if you weren't using this way and doing:

x^2-3=(x+3)^0.5

How would you go about it?
Square both sides to give hence . Then you'd have to solve that by guessing factors. (x-1) and (x+2) are factors that can be obtained here. Then you would divide the polynomial by (x-1) and by (x+2) to get a quadratic, then solve the quadratic to get two more factors.

In total, this will give you four solutions to . However since earlier we squared both sides, some of those solutions would be not what we want, hence we would have to substitute each of the four possible solutions back in to f(x)=f^-1(x) to see if they fit.

As you can see, it's a lot easier to use the f(x)=x method.
5. Thanks.

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