Upon scanning the answer choices, we might recognize that answer choice E, f(x) = 2^x, involves a variable exponent and that the given information that f(a+b) = f (a)f(b) looks A LOT like the Product Law: (k^a)(k^b) = k^(a+b)
So, let's check E first.

E. f(x) = 2^x
If f(x) = 2^x, then f(a) = 2^a, f(b) = 2^b and f(a+b) = 2^(a+b)

We get: f(a)f(b) = (2^a)(2^b)
= 2^(a+b) [apply Product Law]
= f(a+b)
PERFECT!!!

Answer: E

Cheers,
Brent
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Only E satisfies -

\(2^{a+b} = 2^a*2^b\)

Another approach is to test each function to see whether f(a+b) = f(a)f(b)
For example, let's see what happens if a = 1 and b = 1
So, with each function, is it true that f(1 + 1) = f(1)f(1)?
In other words, is it true that f(2) = f(1)f(1)?

A. f(x) = x + 1
Is it true that f(2) = f(1)f(1)?
Plug values into the function to get: 2 + 1 = (1 + 1)(1 + 1)
Simplify: 3 = 4
No good.
ELIMINATE A

B. f(x) = x² + 1
Is it true that f(2) = f(1)f(1)?
Plug values into the function to get: 2² + 1 = (1² + 1)(1² + 1)
Simplify: 4 + 1 = (2)(2)
Simplify: 5 = 4
No good.
ELIMINATE B

C. f(x) = √x
Is it true that f(2) = f(1)f(1)?
Plug values into the function to get: √2 = (√1)(√1)
Simplify: √2 = (1)(1)
No good.
ELIMINATE C

D. f(x) = 1/x
Is it true that f(2) = f(1)f(1)?
Plug values into the function to get: 1/2 = (1/1)(1/1)
Simplify: 1/2 = (1)(1)
No good.
ELIMINATE D

By the process of elimination, the correct answer is E

Cheers,
Brent
_________________

=>

\(A. f(1) = 2, f(2) = 3, f(1)f(2) = 6\), but \(f(1+2) = f(3) = 4\). Choice A is incorrect.
\(B. f(2) = 5, f(3) = 10, f(2)f(3) = 50\), but \(f(2+3) = f(5) = 26\). Choice B is incorrect.
\(C. f(9) = 3, f(16) = 4, f(9)f(16) = 12\), but \(f(9+16) = f(25) = 5\). Choice C is incorrect.
\(D. f(1) = 1, f(2) = \frac{1}{2}, f(1)f(2) = \frac{1}{2}\), but\(f(1+2) = f(3) = \frac{1}{3}\). Choice D is incorrect.
E. Let \(a,b > 0\). Then \(f(a+b) = 2^{a+b} = 2^a2^b = f(a)f(b)\). Choice E is correct.

Therefore, the answer is E.
Answer: E
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