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Home > Documents > Done By : Sara Al-sooj Souhila Sayed Jawaher Al-shaabi Khulood Al-sahlawi Haya Al-naimi.

Done By : Sara Al-sooj Souhila Sayed Jawaher Al-shaabi Khulood Al-sahlawi Haya Al-naimi.

Date post: 21-Dec-2015
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Done By : Sara Al-sooj Souhila Sayed Jawaher Al-shaabi Khulood Al-sahlawi Haya Al-naimi
Transcript

Done By:

Sara Al-soojSouhila Sayed

Jawaher Al-shaabiKhulood Al-sahlawi

Haya Al-naimi

• defenition of exponential function1.Significance <2.Standard Notation.3.Negative Exponents

Exponential function’s Properties (khouloud)

• Graphing the exponential functions (haya and jawaher )

• Exponential equations ( how to solve)• By one-to-one property.By one-to-one property. By finding the x-coordinate in an By finding the x-coordinate in an

exponential function exponential function ( suhaila )

• Conclusion

•The exponential function is one of the most important functions in mathematics. The application of this function to a value x is written as exp(x). Equivalently, this can be written in the form ex, where e is a mathematical constant, the base of the natural logarithm, which equals approximately 2.718281828, and is also known as Euler's number.

•Exponential functions are functions of the form f(x) = bx for a fixed base b which could be any positive real number. Exponential functions are characterized by the fact that their rate of growth is proportional to their value .

One exponential function, f(x)=ex, is distinguished among all exponential functions by the fact that its rate of growth at x is exactly equal to the value ex of the function at x.

•A negative exponent just means that the base is on the wrong side of the fraction line, so you need to flip the base to the other side. For instance, "x –2" just means "x2, but underneath, as in 1/(x2)."

Properties of exponential function

ar as = ar+s

ar / as = ar-s

(ar)s = ars

ar br = (ab)r

Examples:

32 35 = 32+5 = 37

46 /45 = 46-5 = 41

(23)2 = 23*2 = 26

52 32 = (5*3)2 = (15)2 = 225

The general formula of the exponential function is

F(x) = ax

3x

And the base which is a is greater than zero a > 0

the exponent is the "power"

The thing that's being multiplied is

called the "base"

First we will substitute for a and let a (1)

For example when a=3, the function will be f(x) =3x

Therefore to find the points of this function we will have to substitute for x

f(x) =3x

xF(x)

-21/9

-11/3

01

13

29

As x increases with no bound, f(x) increases with no bound

As x decrease with no bound , f(x) reaches to 0

And f(x) = 0 when x approach negative infinity which means that the horizontal asymptote is y=0.

f(x)=5 x

xF(x)

-21/25

-11/5

01

15

225

now we will compare between the two graphs of the functions

As we can see, the greater the value of a the steeper the

curve will be on the y-axis and

the faster it gets to the x-axis from the left.

f)x( =3X

f)x( =5X

So when a (1)

The domain = real numbers (R)

The Range = (0, )

the greater the value of a the steeper the curve will be on the y-axis and the faster it gets to the x-axis from the left.

Now let a (01)

f(x) = 1/3 x

xF(x)

-29

-13

01

11/3

21/9

f(x) = 1/5 x

xF(x)

-225

-15

01

11/3

21/25

As we can see through comparing between the two functions

Domain = real numbers )R(Range= )0, (

As x increase with nobound , f)x( approach 0

As x decrease with no bound , f)x( increase with

no bound

The curve is steeper when the value of a is greater, the

more the graph stretch among the y-axis

f)x( = 1/5 xf)x( = 1/3 x

Now examine the behavior of the when we

multiply it be a negative number

f(x)=1/3 (3 x)k(x)= 3(3 x)m(x)= -3(3 x)

K(x)

m(x)

f(x)

And when we multiply it by -3 the function gets reflected on the x –axis which make the range (- ,0) and as x increases with no bound m(x) decreases with no bound.

• And as x decreases with no bound m(x) approach to zero .

•And when we multiply the function by a negative number the graph gets reflcetwd on the x-axis so the domain is (R) and the range is (- ,0)

•As x increases with no bound , k(x) decreases with no bound

•As x increases with no bound , k(x) approach 0.

Adding or subtracting the function

F(x)= 3 (3 x)+4F(x)=- 3 (3x)-4

As we can see from the graph f(x) is shifted 4 unites upward and g(x) is shifted 4 units downward For f(x):Domain = (R ) Range (4, )As x increases with no bound, f(x) increases with no bound As x decreases with no bound, f(x) approaches to 4 For g(x):Domain = (R ) Range = (-4 )As x increases with no bound, f(x) increases with no bound As x decreases with no bound, f(x) approaches to -4

Methods of solving Methods of solving equationsequations::

By one-to-one property.By one-to-one property.

By finding the x-coordinate By finding the x-coordinate

in an exponential functionin an exponential function..

By Using one-to-one By Using one-to-one propertyproperty::

Examplea) 32x-1=932x-1=32

2x-1=22x=3X=3/2Check:32.3/2-1=32=9.The solution

set is {3/2}

b) 4 |x|=2

)22(|x| =21

22|x|=21

2|x|=1

|x|=1/2

X=±1/2

c) 1/8=4x

2-3=(22)x

2-3=22x

2x=-3

X=-3/2

By finding the x-coordinateBy finding the x-coordinateExampleExample

Let ƒ (x)=2x and (x)=(1/2)1-x Find x if:

a) ƒ (x)=32Because ƒ (x)=2x and ƒ (x)=32,we can find x by solving 2x=322x=322x=25

X=5

b) G (x)=8

Because g (x)=(1/2)1-x and g (x)=8,we can find x by solving(1/2)1-x=8

)1/2(1-x=8)2-1(1-x=23

2x-1=23

X-1=3X=4


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