Question #02485

1 Answer
Oct 12, 2017

See below.

Explanation:

For #a>1 , b>1# and #x>=0#

#ab^x#

Since #b^x# is positive or #1# for all #x >= 0# and #a>1#, we just have a product of positive values, so:

#ab^x>0#

For #a>1 , b>1# and #x<0#

For #x<0# , #b^x= 1/b^x#

So:

#a/b^x>0# (Remember #x# is non negative here)

Taking derivative of #ab^x#

#d/dx(ab^x)= ab^xln(b)#

For zero tangent we would be looking for:

#ab^xln(b)=0=> b^x=0#, but #b^x !=0# ( Not possible for any real positive base )

#lim_(x->oo)(a/b^x)-> 0#

This means that the #x# axis is a horizontal asymptote. The limit to infinity gives 0, but in reality the function just approaches the line #y = 0# it never truly touches it. This can be debatable whether you accept that a tangent of zero exists.