Question

Find the range of ƒ(x) = 2^x + 0.1 + 0.01.

Answer 1

The general form of an exponential function is defined by f (x ) = a ^x-h + k

where a  is positive constant , a  not equals to 1 and a is called base. x is any real number.

The graph has horizontal asympotote of y = k .

Horizontal asympotote y = 0.11

f (x ) = 2^x + 0.1 + 0.01

f (x ) = 2^x + 0.11

y  = 2^x + 0.11

We will find the range by looking at the graph of the exponential function.

Choose values for y and find the corresponding values for x.

x	y = 2x + 0.11	(x, y )
-2	y = 2-2 + 0.11 = 1/4 + 0.11= 0.36	(-2,0.36)
-1	y = 2-1 + 0.11 = 1/2 + 0.11 =0.61	(-1,0.61)
0	y  = 20+ 0.11 = 1 + 0.11 = 1.11	(0,1.11)
1	y = 21 + 0.11 = 2 +0.11 = 2.11	(1,2.11)
2	y = 22 + 0.11 = 4+ 0.11 = 4.11	(2,4.11)

1.Draw a coordinate plane.

2.Plot the coordinate points and dash the horizontal asympotote.

3.Then sketch the graph, connecting the points with a smooth curve.

The range is all positive real numbers (never zero).

Range .

Answer 2

In general, an equation of the form , where a ≠ 0, b > 0, and b ≠ 1, is called an exponential function with base b.

The standard form of an exponential function is .

For , the function is defined for all real values of x. Therefore, the domain is .

The range of is dependent on the sign of a.

To find the range of , start with , since the definition of exponential function.

The function is .

Compare the equation with standard form of an exponential function is .

y = f(x), a = 1 > 0, b = 2 and q = 0.11.

To find the range of , start with .

Add 0.11 to each side of the inequality.

.

The range is .