Question 202899
f(x) = 2f^x-12+30
Is this: {{{f(x) = 2f^(x-12) + 30}}}? If so:<ol><li>It is very unusual (and confusing) for "f" to be used both as the name of the function and as a variable!</li><li>The function "f" is not a function of just "x". It is a function of both "x" and the variable "f"! It would be f(x, f).<li>Please correct the problem and repost it. Please use more parentheses to ensure clarity.</li></ol>
If the second "f" was a typo the function is:
{{{f(x) = 2^(x-12) + 30}}}
then we answer the problem:
(1)The range is (-infinity,infinity).
The range is the set of possible value for the function. So we need to figure out the possible values for {{{2^(x-12) + 30}}}. If we understand exponents then we will realize that:<ul><li>2 to <b>any</b> power can never be zero. Therefore {{{2^(x-12) + 30}}} never have a value of 30</li><li>2 to <b>any</b> power can never be negative. Therefore {{{2^(x-12) + 30}}} can never be less than 30.</li><li>2 to a very large negative power will  be a very tiny fraction which is very close to zero in value. Therefore {{{2^(x-12) + 30}}} can be very, very close to (but never equal to) 30 when x is a large negative number.</li><li>2 to a power will become an infinitely large positive number as x gets to be an infinitely large positive number.</li></ul>Therefore the range is (30, infinity).<br>
(2) The domain is [0,infinity).
The domain is the set of possible x-values. The x in f(x) is found only in the exponent of 2. Since exponents can be <b>any</b> number and since there are no other reasons to exclude x-values (like zeros in denominators, negatives in a square roots, zeros or negatives in logarithms, etc.) the domain of f(x) is all Real numbers: (-infinity, infinity)<br>
(3) There is a vertical asymptote at x=12.
If we had a vertical asymptote at x=12 then 12 would be excluded from the domain. But 12 is in the domain.<br>
(4)There is a horizontal asymptote at y=30.
Horizontal asymptotes occur when the function values approach a certain number when x-values become very large positive or negative numbers. When x is a very large positive number, {{{2^(x-12) + 30}}} becomes a very large positive number. The larger x gets, the larger {{{2^(x-12) + 30}}} gets.
But when x is a very large negative number, {{{2^(x-12)}}} becomes a very tiny fraction and {{{2^(x-12) + 30}}} becomes a number just a tiny bit above 30. The more negative x gets, the tinier the fraction and the closer {{{2^(x-12) + 30}}} gets to 30. So on the left side of the graph of {{{2^(x-12) + 30}}} (where x gets more and more negative, the graph will get closer and closer to y = 30. We do have a horizontal asymptote at y = 30.