Is the function #f(x) = (18)^x# increasing or decreasing?

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Answer 1 · 2015-10-28T14:36:12

depend on the value of #x#. See explanation

if #x -> x>1# then #18^x# is increasing as #x# increases

if #x -> 0 < x <1# then #18^x# is decreasing as #x# approaches 0 from the positive side.

If #x# is negative then we have #1/(18^x)# so:

If #x -> (-1) < x < 0# then as #x# approaches 0 the #18^x# in #1/(18^x)# gets smaller and smaller so #1/18^x# gets bigger and bigger

If #x -> x < (-1) # then the #18^x# in #1/(18^x)# gets bigger and bigger as #x# becomes increasingly les -1. So #1/(18^x)# becomes less and less.

Answer 2 · 2015-10-30T18:44:36

The function is increasing.

An exponential function #y=a^x# is :

increasing for #a>1#
decreasing for #a in (0;1)#
not defined for #a<0# (because negative numbers cannot be raised to rational powers #(-1)^(1/2)=sqrt(-1)# is not a real number)