Thread: Functions

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

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

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

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

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

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

(vi) $\displaystyle 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 $\displaystyle \sqrt{x} $ and $\displaystyle -\sqrt{x} $)

Thanks

Originally Posted by shilz222

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

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

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

are the others correct?

Originally Posted by shilz222

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

Originally Posted by shilz222

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

Yes!

How would you find its inverse?

Originally Posted by shilz222

How would you find its inverse?

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

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

Originally Posted by shilz222

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

Cannot be done easily.

Originally Posted by Plato

Cannot be done easily.

I beg to differ.

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

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

There are no great difficulties here.

-Dan

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