Functions

**shilz222** · Aug 20th 2007, 03:37 PM

Determine which of the following functions $f_i: \mathbb{R} \to \mathbb{R}$ are injective, surjective, and bijective. If they are bijections write down their inverses.

(i) $f_{1}(x) = x-1$ is a bijection (inverse function is $y = x+1$ )

(ii) $f_{2}(x) = x^3$ is a bijection (inverse is $y = \sqrt [3]{x}$ )

(iii) $f_{3}(x) = x^{3}-x$ is not injective, but is surjective.

(iv) $f_{4}(x) = x^{3}-3x^{2}+3x-1$ is not bijective.

(v) $f_{5}(x) = e^{x}$ is bijective (inverse is $y = \ln x$ )

(vi) $f_{6}(x) = \begin{cases} x^{2}\ \text{if}\ x \geq 0 \\ -x^{2}\ \text{if}\ x \leq 0 \end{cases}$ is bijective (inverse is same conditions with function as $\sqrt{x}$ and $-\sqrt{x}$ )

Thanks

**Plato** · Aug 20th 2007, 03:49 PM

Originally Posted by shilz222

Determine which of the following functions ${\color{red} f_i: \mathbb{R} \to \mathbb{R}}$ are injective, surjective, and bijective. If they are bijections write down their inverses.
(v) $f_{5}(x) = e^{x}$ is bijective (inverse is $y = \ln x$ )

You do know that $e^x > 0$ ? Can that be onto?

**shilz222** · Aug 20th 2007, 03:52 PM

arrgh, idiot I am. It would have an inverse if it was defined on $R \to R^{+}$ right?

**shilz222** · Aug 20th 2007, 09:48 PM

are the others correct?

**topsquark** · Aug 21st 2007, 06:14 AM

Originally Posted by shilz222

Determine which of the following functions $f_i: \mathbb{R} \to \mathbb{R}$ are injective, surjective, and bijective. If they are bijections write down their inverses.
(iv) $f_{4}(x) = x^{3}-3x^{2}+3x-1$ is not bijective.

Ignoring any polynomial math that you ought to recognize, have you looked at the graph of this function? It looks to me like it has an inverse...

-Dan

**Plato** · Aug 21st 2007, 07:37 AM

Originally Posted by shilz222

arrgh, idiot I am. It would have an inverse if it was defined on $R \to R^{+}$ right?

Yes!

**shilz222** · Aug 21st 2007, 10:58 AM

How would you find its inverse?

**Plato** · Aug 21st 2007, 11:39 AM

Originally Posted by shilz222

How would you find its inverse?

You did above: $y=\ln (x)$ .

**shilz222** · Aug 21st 2007, 11:45 AM

the one Topsquark was talking about (the graph). Can you find the inverse easily?

**Plato** · Aug 21st 2007, 01:22 PM

Originally Posted by shilz222

the one Topsquark was talking about (the graph). Can you find the inverse easily?

Cannot be done easily.

**topsquark** · Aug 21st 2007, 02:22 PM

Originally Posted by Plato

Cannot be done easily.

I beg to differ.

$f(x) = x^3 - 3x^2 + 3x - 1 = (x - 1)^3$

So the inverse function is
$f^{-1}(x) = \sqrt[3]{x} + 1$

There are no great difficulties here.

-Dan

**Plato** · Aug 21st 2007, 02:47 PM

I really missed that one.
I had in mind a general qubic that is one-to-one.

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