Functions

• Aug 20th 2007, 04:37 PM
shilz222
Functions
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
• Aug 20th 2007, 04:49 PM
Plato
Quote:

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?
• Aug 20th 2007, 04:52 PM
shilz222
arrgh, idiot I am. It would have an inverse if it was defined on $R \to R^{+}$ right?
• Aug 20th 2007, 10:48 PM
shilz222
are the others correct?
• Aug 21st 2007, 07:14 AM
topsquark
Quote:

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
• Aug 21st 2007, 08:37 AM
Plato
Quote:

Originally Posted by shilz222
arrgh, idiot I am. It would have an inverse if it was defined on $R \to R^{+}$ right?

Yes!
• Aug 21st 2007, 11:58 AM
shilz222
How would you find its inverse?
• Aug 21st 2007, 12:39 PM
Plato
Quote:

Originally Posted by shilz222
How would you find its inverse?

You did above: $y=\ln (x)$.
• Aug 21st 2007, 12:45 PM
shilz222
the one Topsquark was talking about (the graph). Can you find the inverse easily?
• Aug 21st 2007, 02:22 PM
Plato
Quote:

Originally Posted by shilz222
the one Topsquark was talking about (the graph). Can you find the inverse easily?

Cannot be done easily.
• Aug 21st 2007, 03:22 PM
topsquark
Quote:

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
• Aug 21st 2007, 03:47 PM
Plato
I really missed that one.
I had in mind a general qubic that is one-to-one.