Prove $f(x) = 3^x$ is not onto.

Question

Give a counterexample to prove $f(x) = 3^x$ is not onto.

A function is onto if for all $y$ in the codomain, there exists an $x$ in the domain such that $f(x) = y$. Essentially, the range of our function $f$ is equal to the codomain.

I know that $f(x)=3^x$ is not onto, but I'm having trouble finding a counterexample to prove this.

We know the domain is the real numbers and the codomain is the positive real numbers. So we want to find a positive real number that doesn't equal $3^x$, correct?

Specify the domain and range of $f$. – creative Sep 08 '14 at 10:04 — creative, Sep 08 '14 at 10:04

score 3 · Answer 1 · answered Sep 08 '14 at 09:23

3

Incorrect. A positive real number that does not equal $3^x$ for any $x$ does not exist.

$f(x) = 3^x$, viewed as a function from $\mathbb R$ to $\mathbb (0,\infty)$ IS onto. Every positive real number can be attained as $3^x$ for some real $x$.

$f(x)=3^x$ as a function from $\mathbb R$ to $\mathbb R$ is NOT onto. You can probably see what real numbers cannot be reached, since you already know that for each $x$, $3^x\in(0,\infty)$.

answered Sep 08 '14 at 09:23

5xum

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The entire question was to prove that for all X, there exists no function f: X -> 3^x such that f is not onto. – user174502 Sep 08 '14 at 12:57
Well, for every $X$, the function $f$ is onto if the codomain if $f(X)$. – 5xum Sep 08 '14 at 13:06

score 2 · Answer 2 · answered Sep 08 '14 at 09:47

The question "Prove $f(x)=3^x$ is not onto" is not a complete question. You have to know the sets $A$ and $B$ such that $f: A\rightarrow B$. It is reasonable, though, to expect that the intended question was:

Give a counter-example to prove that $$f: \mathbb{R}\rightarrow\mathbb{R} $$ $$f(x) = 3^x $$ is not onto. For this it is easy to find a $y\in\mathbb{R}$ such that $f(x)=y$ is impossible for $x\in\mathbb{R}$.

Let $y = -1$ then there exists no $x\in\mathbb{R}$ such that $f(x)=3^x=-1$. This is a counter-example and we are done.

The entire question was to prove that for all X, there exists no function f: X -> 3^x such that f is not onto. — user174502, Sep 08 '14 at 12:57

score 1 · Answer 3 · answered Sep 08 '14 at 10:17

If you take $f:\mathbb{R}\to \mathbb{R}$, then $f(x)=3^x$ is not onto for a simple reason.

Defn : A function $f:A\to B$ is onto if B=ran(f), and

$ran(f)=\{b\in B:\exists a \in A \ni f(a)=b\}$.

Now $3^x\ne 0, \forall x\in \mathbb{R}$, which means $\nexists x \in \mathbb{R}$ for which $3^x=0$, so B $\ne ran(f)$
f is thus not onto. $\blacksquare$

score 0 · Answer 4 · answered Sep 08 '14 at 09:39

Proof that

$$f:\Bbb R\to (0,\infty)\;,\;\;\;f(x):=3^x\;,\;\;\;\text{is onto}:$$

$$\forall y\in (0,\infty)\;,\;\;3^x=y\iff x=\log_3y=\frac{\log y}{\log 3}$$

and we're done as we know the (napierian or whatever) logarithm is defined on (all) the positive reals and $\;\log 3\neq 0\;$

Prove $f(x) = 3^x$ is not onto.

4 Answers4