Question

What is the range of the exponential function shown below? F(x)=11×(1/3)^x

Answer 1

$(0, \infty)$

Answer 2

Answer:

The range of the given exponential function is (0,∞).

Step-by-step explanation:

The general form of an exponential function is

$y=ab^x$

Where, a is initial value, b>1 is growth factor and 0<b<1 is decay factor.

The given function is

$F(x)=11\times (\frac{1}{3})^x$

Here $a=11,b=\frac{1}{3}$. It means the initial value of the function is 11 and the decay factor is $\frac{1}{3}$. The value of $(\frac{1}{3})^x$ is positive for all real values of x.

$(\frac{1}{3})^x>0$

Multiply both sides by 11.

$11(\frac{1}{3})^x>0$

$F(x)>0$

It means the value of F(x) is positive for all real values of x.

Range is the set of outputs. So, the range of the function all positive real number except 0.

Range = (0,∞)

Therefore the range of the given exponential function is (0,∞).