Question

Find the range of f(x) = -square roots of 3x+9 -1

Answer 1

The function \[f(x)=-\sqrt{3x+9}-1\]looks quite complicated.
Could you anser the question for \[g(x)=\sqrt{x}\]?

Answer 2

If you can, you're already halfway understanding the range of f!

Answer 3

Not quite

Answer 4

I am having lots of trouble understanding the ranges

Answer 5

OK, here is the graph of my simpler function g:

In √x, you only can input numbers that are 0 or larger, because the root of a negative number doesn't exist (or: is not a real number).
This means: the domain of √x is real all numbers that are 0 or larger.
This why you see no graph left of the y-axis btw.

The range consists of all outcomes of √x. They are also 0 or larger. You can see that in the graph: as x grows larger, g(x) grows larger also, only not so fast.

So: the range of √x is all real numbers equal of larger than 0.

Answer 6

Now suppose you have \[h(x)=-\sqrt{x}\]Here everything is the same as in g(x), only after the calculation of the root you put a minus in front of the outcome.
This means all the positive outcomes become negative now.

So the range of h are all real numbers equal or *smaller* than 0.
Written as an interval: (-∞,0]. See updated graph.

Do you understand so far?

Answer 7

@cassandre: do you copy?

Answer 8

yes i get it so far

Answer 9

sorry, i took a while to answer but i really do need help with this chapter

Answer 10

OK, so now see this new function k(x) = −√x - 1.

This is h(x), only now 1 less. See graph to see what this means.

Because all the outcomes are now 1 less, it is easy to see that the range of k is (-∞,-1].
This is also the range of your function! Reason: the same changes have been made.

The only thing that is different in function f is the number under the root sign.
It has no effect on the range, because, just like x it can *in principle* take on any value.
This means it's you responsibility to choose x in such a way that 3x+9 is equal or larger than 0.