How do you identify equations as exponential growth, exponential decay, linear growth or linear decay #y=(3/4)^3#?

1 Answer
Jul 28, 2015

Answer:

If the function is #y=f(x)=(3/4)^{x}#, then it's exponential decay.

Explanation:

Functions of the form #y=f(x)=a*b^{x}#, where #a>0#, #b>0#, and #b!=1# are called exponential functions. If #0<b<1#, then it's a decreasing function and is called an exponential decay function. If #b>1#, then it's an increasing function and is called an exponential growth function.

Exponential decay functions have the property that each 1 unit increase in #x# results in a decrease of #y# by #100*(1-b)# percent. For example, if #b=0.96# and #x# increases by 1 unit, then #y# goes down by 4%. Here are some algebraic details that confirm this: #(Delta y)/y=(a*b^(x+Delta x))/(a*b^{x})=(cancel(a)*cancel(b^{x})*b^{Delta x})/(cancel(a)*cancel(b^{x}))=b^{Delta x}=b^{1}=0.96#, a 4% decrease.

On the other hand, exponential growth functions have the property that each 1 unit increase in #x# results in an increase of #y# by #100*(b-1)# percent. For example, if #b=1.08# and #x# increases by 1 unit, then #y# goes up by 8%. Here are some algebraic details that confirm this: #(Delta y)/y=(a*b^(x+Delta x))/(a*b^{x})=(cancel(a)*cancel(b^{x})*b^{Delta x})/(cancel(a)*cancel(b^{x}))=b^{Delta x}=b^{1}=1.08#, an 8% increase.

You can watch my "Quick Precalc Review" videos to see a lot of examples like this (I start by talking about linear functions but also get to exponential functions by Video #7).

Quick Precalculus Review for Calculus, Video #1