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Done by: Wa’ad Qayed Al-Johara Al-Thani Moza Salam Zahra Abdulla For Dr.Fuad Almuhanadi

Exponential Function. Done by: Wa’ad Qayed Al-Johara Al-Thani Moza Salam Zahra Abdulla For Dr.Fuad Almuhanadi. Exponents: Basic Rules (zahra+Wa’ad) Exponential Equations. (Moza) Exponential function and their graphs. (Moza+ ALjohara) Application on Exponential function. (Wa’ad).

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Done by: Wa’ad Qayed Al-Johara Al-Thani Moza Salam Zahra Abdulla For Dr.Fuad Almuhanadi

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  1. Exponential Function Done by: Wa’ad Qayed Al-Johara Al-Thani Moza Salam Zahra Abdulla For Dr.Fuad Almuhanadi

  2. Exponents: Basic Rules(zahra+Wa’ad) • Exponential Equations. (Moza) • Exponential function and their graphs. (Moza+ ALjohara) • Application on Exponential function.(Wa’ad)

  3. Section: 1 Exponents: Basic Rules

  4. the exponent is the "power" The thing that's being multiplied is called the "base" Exponents are shorthand for multiplication process. The "exponent" stands for how many times the thing is being multiplied. For ex. (5)(5)(5)= 125 this the same as 53. 53

  5. Basic rules 1- Simplify (x3)(x4): (x3)( x4) = (xxx)(xxxx) ( xm ) (xn) = x(m+n) = xxxxxxx = x7 So, (x3)( x4) = (x3+4 ) =x7 However we can’t simplify (x4)(y3)= xxxxyyy = (x4)(y3).

  6. (x2)4 = (x2)(x2)(x2)(x2) • 2-Simplify (x2) 4 = (xx)(xx)(xx)(xx) (xm)n=xmn   = xxxxxxxx   = x8 (x2) 4= x(2.4)= x8 So,

  7. 1- (xy 2) 3 = (xy 2)(xy 2)(xy 2) = (xxx)(y 2 y 2 y 2) = (xxx)(yyyyyy) = x 3y 6 = (x) 3 (y 2) 3 This will only applied for multiplication and division. However we cannot apply it for addition and subtraction. Solve (xy2)3 2-

  8. For example (3+2)2 If we distribute the power 2 then, (3 ) 2 +(2) 2 = 9+4 =13 And if we not distributethe power 2 then, (3+2) 2 = (5) 2 = 25 = 13

  9. Example (x – 2) 2 (x-2)2 = (x – 2) (x – 2) = xx – 2x – 2x + 4 = x2– 4x + 4

  10. 35 = 36÷ 3 = 243 34 = 35 ÷ 3 = 81 33 = 34 ÷ 3 = 27 32 = 33 ÷ 3 = 9 31 = 32 ÷ 3 = 3 Then logically 30 = 31 ÷ 3 = 3 ÷ 3 = 1. Anything to the power zero is just "1". (3)0= 1 a0 =1 3- [(3x4y7z12)5 (–5x9y3z4)2]0 =1 m0 = m(n-n) = mn × m-n = mn ÷ mn = 1

  11. Dealing with negative exponent

  12. negative exponent • 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.

  13. Write x-4 using only positive exponents. Write x2 / x-3 using only positive exponents.

  14. Write 2x-1using only positive exponents Note that the "2" above does not move with the variable, the exponent is only on the "x".

  15. Write (3x)-2 using only positive exponents. Write (x -2/ y -3) -2 using only positive exponents

  16. The radical The radical of any number can express in exponent. or

  17. 2- Exponential Equations

  18. 1- Solve 5x = 53. x=3 2-Solve 32x–1 = 27. 32x–1 = 27 32x–1 = 332x – 1 = 32x = 4 x = 2

  19. 3- Solve 3x2–3x = 81. 3x2–3x = 81 3x2–3x = 34x2 – 3x = 4 x2 – 3x – 4 = 0 (x – 4)(x + 1) = 0 x = –1, 4 4- Solve 42x2+2x = 8. 4 = 22 8 = 23 42x2+2x = (22)2x2+2x = 24x2+4x Now I can solve: 42x2+2x = 8 24x2+4x= 23 4x2 + 4x = 3 4x2 + 4x – 3 = 0 (2x – 1)(2x + 3) = 0 x = 1/2 , –3/2

  20. 5- Solve 4x+1 = 1/64. 4x+1 = 1/64 4x+1 = 4–3x + 1 = –3x = –4

  21. 3- Exponential function and their graphs The general formula of the exponential function is: F(x) = ax

  22. let a  (01)  (1) where a is the base such that a > 0 and a not equal to 1. As an example will take a=2 so, the function will be f(x) = 2x To find the point of the function we will substitute in (x)with different points.

  23. f(x) = 2x y 1 x

  24. As x increases with no bound, f(x) increases with no bound Lim f (x) = positive infinity when (x) increase to positive infinity. • As x decreases with no bound , f(x) approaches 0 Lim g (x) = 0 when (x) decrease to negative infinity. • This means that the line y=0 works as horizontal asymptote.

  25. the value of the graph at zero will equalone (f (0) =2(0)=1). • The domain of this graph is equal to thereal numbers(R) • the range of this graph is(0,)

  26. g(x) = 3x y 1 x

  27. Comparison between f(x) and g(x) F(x)=2x G(x)=3x y 1 x

  28. In the case ofa  (01)  (1) • The domain of the function f(x)=axis the real numbers (R). • The range of the function f(x)=axis the positive numbers between (0, ) • The greater is a the more rapidly the curve of the function increases (the more steeply) on the right of the y- axis and the faster it approaches the x-axis on the left of the y-axis. Therefore, the graph stretches more among the x-axis.

  29. y • Anther example of exponential function: d(x) =1/2(2) x and g(x) = 2(2)x G(x)=2(2)x 2 D(x)=1/2(2)x F(x)=(2)x 1 x the y-axis of the function g(x) = 2(2 )x multiplies by 2 and the function d(x) = 1/2(2 )x multiplies by half(1/2) . This appears when we find g(0)=2 and f(0)=1 we can see that it is multiply by 2 and the function stretch among the y-axis. And d(0)=1/2 and f(0)=1, the function shrink by 1/2 among the y-axis .

  30. f(x) = (2 )^x g(x) = -(2 )^x If we multiply by a negative number:

  31. range will be (0,-) and the domain equal the real number • As x increases with no bound, g(x) decreases with no bound Lim g (x) = positive infinity when (x) decrease to negative infinity. • As x decreases with no bound , g(x) approaches 0 Lim g (x) = 0 when (x) decrease to negative infinity. • This means that the line y=0 works as horizontal asymptote.

  32. The third example: g(x) = 2(2 )x+3 and d(x) = 1/2(2 )x-3 g(x) = 2(2 ) x +3 d(x) = 2(2 ) x -3

  33. g(x) = 2(2 )x +3 • As x increases with no bound, g(x) increases with no bound • Lim g (x) = positive infinity when (x) increase to positive infinity. • As x decreases with no bound , g(x) approaches 3 • Lim g (x) = 3 when (x) decrease to negative infinity. • This means that the line y=3 works as horizontal asymptote. • The domain is the real number, but the range is equal to (3, )

  34. d(x) = 2(2 )x-3 • As x increases with no bound, d(x) increases with no bound • Lim g (x) = positive infinity when (x) increase to positive infinity. • As x decreases with no bound , d(x) approaches -3 • Lim d (x) = -3 when (x) decrease to negative infinity. • This means that the line y=-3 works as horizontal asymptote. • The domain is the real number, but the range is equal to (-3, )

  35. That the value of the graph at zero will equal one (g (0) =2 0=1). The domain of this graph is equal to the real numbers (R), and the range of this graph is (0,). • As x increases with no bound, f(x) approaches 0 • That is limf(x) =0, when (x) increases to positive infinity. • As x decreases with no bound, f(x) increases with no bound • This means that the line y=0 works as horizontal • The fourth example: g(x)=2(-x) y f(x)=2x g(x)=2(-x) 1 x

  36. Example: let a  (01) f(x) =1/2x y 1 x

  37. g(x) = 1/3x y 1 x

  38. As x increases with no bound, g(x) approaches 0 • That is limg(x) =0, when (x) increases to positive infinity. • As x decreases with no bound, g(x) increases with no bound • That is limg(x) =+∞ , when (x) decreases to negative infinity. • This means that the line y=0 works as horizontal • The greater is a the more rapidly the curve of the function increases ( the more steeply ). The more the graph stretch among the y-axis. y G(x)=(1/3)x F(x)=(1/2)x 1 x

  39. g(x) = 2(2)^(x+3) f(x) = 2(2)^(x) If we added or subtracted from (X). To compare this is the example: function g(x) = 2(2) ^(x+3),f(x) = 2(2) ^x and d(x) = 2(2) ^(x-3) d(x) = 2(2)^(x-3)

  40. Application of Exponential function

  41. On this application we will study the behavior of the antibiotic drugs' on the body every period of time. The behavior of the drugs on the body is the same as the behavior of the exponential function. The following graph shows the behavior of the drug during the first 10 hours.

  42. On the pervious graph we see that the amount of the drug among ten hours decline to 1.5. we use this graph to find the equation of the line .

  43. We use the general formula: We chose the point (0, 10) to find first the value of a by substation on the pervious equation to be:

  44. By this equation we find the value of a to be 10 and it will be constant for this values. Now to find the value of be we will substitute on the equation to find the value and then find the average. By this the equation of the graph will be:

  45. We apply the function to the amount of the drug on the blood for one day for each six hour

  46. This is the values of the drugs for two days

  47. This is the values of the drugs for four days

  48. Observation: We can see that the value of the drugs on the bloodstream on the body remains constant and it will continue its maintaining and the patient stops taking the drug. So that way the doctors always told us to take the antibiotics after specific time.

  49. Thank you

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