Question

the exponential function ( f ) is defined by ( f(x) = ab^x ), where ( a ) and ( b ) are positive constants. the table gives values of ( f(x) ) at selected values of ( x ).

( x )	( f(x) )
1	( \frac{3}{2} )
2	3
3	6
4	12

which of the following statements is true?

a ( f ) demonstrates exponential decay because ( a > 0 ) and ( 0 < b < 1 )

b ( f ) demonstrates exponential decay because ( a > 0 ) and ( b > 1 )

c ( f ) demonstrates exponential growth because ( a > 0 ) and ( 0 < b < 1 )

d ( f ) demonstrates exponential growth because ( a > 0 ) and ( b > 1 )

Explanation:

Step1: Find the value of \(a\)

Given the exponential function \(f(x)=ab^{x}\), when \(x = 0\), \(f(0)=\frac{3}{4}\). Substitute \(x = 0\) into the function: \(f(0)=ab^{0}\). Since \(b^{0}=1\) for any \(b>0\), we have \(a=\frac{3}{4}\).

Step2: Find the value of \(b\)

Now we know \(a = \frac{3}{4}\), and we can use another point to find \(b\). Let's use \(x = 1\), where \(f(1)=\frac{3}{2}\). Substitute into the function: \(\frac{3}{2}=\frac{3}{4}\times b^{1}\). Solve for \(b\): \(b=\frac{\frac{3}{2}}{\frac{3}{4}}=\frac{3}{2}\times\frac{4}{3}=2\). Since \(b = 2>1\) and \(a=\frac{3}{4}>0\), the function demonstrates exponential growth (exponential growth occurs when \(a>0\) and \(b > 1\) in \(f(x)=ab^{x}\)).

Step3: Analyze the options

• Option A: Says exponential decay with \(0 < b<1\), but we found \(b = 2>1\), so A is wrong.
• Option B: Says exponential decay with \(b>1\), but decay requires \(0 < b<1\), so B is wrong.
• Option C: Says exponential growth with \(0 < b<1\), but \(b = 2>1\), so C is wrong.
• Option D: Says exponential growth with \(a>0\) and \(b>1\), which matches our findings (\(a=\frac{3}{4}>0\) and \(b = 2>1\)).

Answer:

D. \(f\) demonstrates exponential growth because \(a>0\) and \(b > 1\)