# Math Help - Functions

1. ## 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

2. 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?

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

4. are the others correct?

5. 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

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

7. How would you find its inverse?

8. Originally Posted by shilz222
How would you find its inverse?
You did above: $y=\ln (x)$.

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

10. Originally Posted by shilz222
the one Topsquark was talking about (the graph). Can you find the inverse easily?
Cannot be done easily.

11. 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

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