Question

Which of these values for p and a will cause the function f(x)=Pax to be an exponential growth function

Answer 1

The option are missing in the question. The options are :

A.  P = 2, a = 1

B.  $P=\frac{1}{2} ; a =\frac{1}{3}$

C. $P=\frac{1}{2} ; a =1$

D. P = 2, a = 3

Solution :

The given function is $f(x)= Pa^x$

So for the function to be an exponential growth, a should be a positive number and should be larger than 1. If it less than 1 or a fraction, then it is a decay. If the value of a is negative, then it would be between positive and negative alternately.

When the four option being substituted in the function, we get

A). It is a constant function since $2(1^x)=2$

B). Here, the value of a is a fraction which is less than 1, so it is a decay function. $f(x)=\frac{1}{2}\left(\frac{1}{3}\right)^x$

C). It is a constant function since the value of a is 1.

D). Here a = 3. So substituting, as the value of x increases by 1, the value of the function, f(x) increases by 3 times.

  $f(x)=2(3)^x$

Therefore, option (D). represents an exponential function.