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The function. Objectives: To recognise the number e and be able to differentiate y=e kx To recognise the inverse of y=e kx. Let us consider these graphs. We notice that all graphs pass through the same point (0,1) This is because anything to the power zero gives the answer 1.
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The function Objectives: • To recognise the number e and be able to differentiate y=ekx • To recognise the inverse of y=ekx
Let us consider these graphs. We notice that all graphs pass through the same point (0,1) This is because anything to the power zero gives the answer 1. From the graphs we can see that as the value of the base increases, the gradient increases.
The gradient of the graph of at x = 0 is 0.693. The gradient of the graph of at x = 0 is 1.098 There must then be some value between 2 and 3 which gives us a gradient of 1 when x = 0 . This value is 2.71828... It continues on irrationally. This number is a special function and is known as the Exponential function and is denoted by the letter e.
The value of a where the is an irrational number, written as e,where gradient ofequals
The Inverse of is a one-to-one function so has an inverse function. N.B. The domain is . We can sketch the inverse by reflecting in y = x.
is a growth function. • At every point on , the gradient equals y: • (3 d.p.) • The inverse of is is defined for x > 0 only SUMMARY ( log with base e )
Calculus, ex and logarithms Objectives: To apply the laws of logs to solve equations involving ex and lnx To be able differentiate and integrate functions involving exponential functions and natural logarithms.
Activities: • Multiple Choice • Natural Log Examples • Tarsia puzzle
Homework Ex. 4A: 1,2,3 (2 parts of each) & 6 Ex. 4B: 1,2,3 (2 parts of each) & 5 Ex 4C: choose 3 (3 parts of each) Ex 4E: 1,2,3,6 & one other Mixed Ex: 2, 3 & choose 1 exam q. [A]