# Find domain and codomain of this function…

So, the function is: $$z(x)=\sqrt{-4^x+6\cdot2^x-8}$$ We have to find domain and codomain. There are many more function in the exercise, I just want to know how it's done, so I can do the next examples on my own.

$$-4^x+6\cdot2^x-8=-\{(2^x)^2-2\cdot2^x\cdot3+3^2\}+1=1-(2^x-3)^2$$

We know for real $a,a^2\ge0$

$\implies (2^x-3)^2\ge0\iff -(2^x-3)^2\le0\iff 1-(2^x-3)^2\le1$

$\implies z(x)\le1$

For real $z(x),$

$-4^x+6\cdot2^x-8\ge0\iff 1-(2^x-3)^2\ge0\iff (2^x-3)^2\le1$ $\iff-1\le 2^x-3\le1 \iff2\le 2^x\le4\iff1\le x\le2$

To find the domain solve the follwoing exponential equation $-4^x + 6 \cdot 2^x - 8 \ge 0$ and the codomain is $x \in \mathbb{R}^{+} \cup \{0\}$, because square roots give only positive numbers.

• A codomain is $\Bbb R^+\cup\{0\},$ certainly, but $\Bbb R$ also works, and in fact, any superset of $[0,1]$ does the trick. – Cameron Buie Sep 22 '13 at 16:39
• I think that negative numbers don't work, because $z(x)$ can have only negative because the RHS can't have negative values. – Stefan4024 Sep 22 '13 at 16:45
• Negative numbers won't be elements of the range, that's true. A codomain just has to be a superset of the range, though. – Cameron Buie Sep 22 '13 at 16:46
• You're right I messed the terms range and codomain. – Stefan4024 Sep 22 '13 at 16:54
• Even for range, have you checked if the content of the square root gives every real number from 0 to $+\infty$? – peterwhy Sep 22 '13 at 16:58

Domain I assume you are talking about this function for real numbers. Then the function is only defined if the content inside the square root is non-negative:

$$-4^x + 6\cdot2^x - 8 \ge 0$$

I guess you can solve for the set of $x$ that fits in this inequality. This gives the largest domain.

Codomain If you want the smallest codomain, you should find the range of $\sqrt{-4^x + 6\cdot2^x - 8}$, by finding the range of $-4^x + 6\cdot2^x - 8$, by finding the range of $-4^x + 6\cdot2^x$, by finding the range of $-\left(2^{x}\right)^2+6\cdot2^x$, for the domain found above.