Question 256549
(a) f(g(x)) will become 2^cx. As x increases f(g(x)) increases
(b) f(g(x)) will become 2^c/x. As x increases f(g(x)) decreases
(c) f(g(x)) will become 2^x/c. As x increases f(g(x)) increases
(d) f(g(x)) will become 2^(x-c). As x increases f(g(x)) increases
(e) f(g(x)) will become 2^(log_c(x)). As x increases f(g(x)) increases
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Now (b) decreases, so it is out of contention.
Since c is a constant, let c = 3. We get
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(a) f(g(x)) will become 2^(3x). As x increases f(g(x)) increases
(c) f(g(x)) will become 2^(x/3). As x increases f(g(x)) increases
(d) f(g(x)) will become 2^(x-3). As x increases f(g(x)) increases
(e) f(g(x)) will become 2^(log_c(x)). As x increases f(g(x)) increases
Now, (c) < (a), so (c) is out. (d) < (a) so (d) is out. We now have:
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(a) f(g(x)) will become 2^(3x). As x increases f(g(x)) increases
(e) f(g(x)) will become 2^(log_c(x)). As x increases f(g(x)) increases
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It turns out that (a) grows a lot faster than (e) so (a) is the largest value.