How is the range of y = 3^x - 4 ..?
(-4, inf)
How is the range of the previously stated equation (-4, infinity)? Shouldn't it be (-3, infinity)? Since the graph of y = 3^x is moved down 4 units and y = 3^x would equal 1 when x = 0 (since all numbers to the 0th power equal 1). 1 minus 4 equals -3... not -4
Right, thanks :) I'm mostly wondering WHY/HOW that is...
well , what is the range of 3^x?
(1, inf) right?
no, it's (0,inf)
3^0 = 1
How is that?
if x approaches negative infinity, y approaches 0+
so numbers like 0.000001 is in the range, yes?
How could the value of y ever equal less than 1 if x^0 = 1... and the exponents can't be negative
I was under the impression the exponents couldn't be negative for some reason... maybe not?
yes, exponent CAN be negative. for example, 3^(-2) = 1/3^2 = 1/9
so do you see why the range of 3^x is (0,inf)? and since 3^x - 4 is just 4 units down, the range is (-4,inf)
Ohhh.. gotcha. So the domain of a^x is (-inf, inf) and the range is (0, inf)?
Why is the range always positive?
because the values of "a" are defined to be 0 < a < 1, or a > 1.
i.e is is positive. A positive number raised to any power is always positive
Oh okay, so the base cannot be negative?
Got it! Thanks for clearing that up
well, base can certainly be negative but there are many problems for example (-3)^(1/3) is defined, but (-3)^(1/2) isn't.
Hmmm
But it's still always positive
the problem is if the base is negative, the function is no longer continuous . What what exponent functions to be continuous
no (-3)^(1/3) isn't positive
anyways, it has something to do with continuity. But that's higher math. For now, just now that the base is always positive, except for 1
Thanks a bunch :) By "higher math", what do you mean? Calculus?
yes, calculus or higher
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