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

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

asked Jun 11, 2014 in PRECALCULUS by bilqis Pupil

2 Answers

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

where 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 .

f (x ) = 2x + 0.1 + 0.01

f (x ) = 2x + 0.11

y  = 2x + 0.11

Horizontal asympotote y = 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.

graph the equation x=y^2

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

Range image.

 

answered Jun 11, 2014 by david Expert
0 votes

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

The standard form of an exponential function is image.

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

The range of image is dependent on the sign of a.

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

The function is image.

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

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

To find the range of image , start with image.

Add 0.11 to each side of the inequality.

image

image.

The range is image.

answered Jun 11, 2014 by casacop Expert

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