Find each function value, if it exists for f(t) = sqrt t^2+1.
f(-1)?

Question

across · Answer

Tell me, what is confusing you about this problem?

anonymous · Answer

Well I know there is no square root of a negative number, right?

anonymous · Answer

So I came up with zero as my answer but it says that is not correct.

across · Answer

I will assume the function is the following:$$f(t)=\sqrt{t^2+1}.$$What happens if you pass it a -1? I.e.,$$f(-1)=\sqrt{(-1)^2+1}?$$

across · Answer

By the way, you're right that you can't take the square root of a negative number (that is, if you're not working with complex numbers).

anonymous · Answer

My other option is that the square root is not a real number and that isn't right either, so I am stumped.

across · Answer

$$f(-1)=\sqrt{(-1)^2+1}=\sqrt{1+1}=\sqrt{2}.$$ Do you agree?

anonymous · Answer

How did you get (-1)^2 = 1?  I figured it as being -2

across · Answer

Because$$(-1)^2=(-1)(-1)=1.$$A negative number times a negative number is always a positive number.

anonymous · Answer

okay, well that is where I am messing up.

anonymous · Answer

I also forgot to put the "sqrt" symbol in front of my answer.  That was stupid!

across · Answer

No, that's fine; we all forget a thing or two at times. ;]

anonymous · Answer

Thanks for your help!