Can someone please describe the end behavior of the function?

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2 Answers
RedRobin9688
Jun 26, 2018

As #x# increases #f(x)# approaches #-2#

Explanation:

#lim_(x->oo) k^x ->0# when #k<1#

Therefore

#lim_(x->oo) (2/3)^x-2 rarr 0-2=-2#

SCooke · Shwetank Mauria
Jun 26, 2018

As x increases, #f(x)# approaches #-2#.

Explanation:

From a simple algebraic perspective, just note that any number larger than one (1) raised to any positive exponent will increase to infinity. Any number LESS than one will rapidly decrease - approaching zero (0), if not quite ever actually reaching it.

Thus, we can see that the first term, #(2/3)^x# , is going to approach #0# as x increases. Thus the final result of f(x) will be approaching #-2#.

A couple of quick calculations will show that trend very simply.

For x = 0
#f(x) = (2/3)^0 - 2 = 1 - 2 = -1#
For x = 1
#f(x) = (2/3)^1 - 2 = (2/3) - 2 = -1.33#
For x = 10
#f(x) = (2/3)^10 - 2 = (0.017) - 2 = -1.983#
For x = 100
#f(x) = (2/3)^100 - 2 = (2.435 xx10^(-18)) - 2 = -2.0000...#

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