MAM
Functions+Notes+_updated_.pdf

# For example the value of an investment bearing

• Notes
• 58

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proportional to their size. For example, the value of an investment bearing interest that is continuously compounding increases at a rate proportional to that value. The mass of undecayed radioactive material in a sample decreases at a rate proportional to that mass. All of these phenomena, and others exhibiting similar behavior, can be modelled mathematically in the same way. It Q = Q ( t ) denotes the value of a quantity Q at time t , and if Q changes at a rate proportional to its size, then dQ dt = kQ, where k is the constant of proportionality. The equation above is called the differential equation of exponential growth or decay because, for any value of the constant C , the function Q ( t ) = Ce kt satisfies the equation. Moreover, the initial-value problem dQ dt = kQ, Q (0) = Q 0 has unique solution Q ( t ) = Q 0 e kt . We say that the quantity Q exhibits exponential growth if k > 0 and exponential decay if k < 0. Doubling Time. Half-life Exponential growth is characterized by a fixed doubling time . If t d is the time at which Q has double from its size at t = 0, then 2 Q (0) = Q ( t d ) = Q 0 e kt d = Q (0) e kt d , since Q (0) = Q 0 . But then e kt d = 2, which implies that kt d = ln 2, or t d = ln 2 /k . Similarly, exponential decay involves a fixed halving time, usually called the half-life .

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2. Functions 29 Application. How long, to the nearest year, will it take an invest- ment to double its value if invested at 6% annually compounded continuously? Solution: Let us assume that the amount invested is P . Then recall that the future value A ( t ) of P invested at 6% p. a. compounded continuously is given by the following exponential growth function: A ( t ) = Pe 0 . 06 t We then have to find T such that A ( T ) = 2 A (0) = 2 P . Since A ( T ) = Pe 0 . 06 T we obtain the equation Pe 0 . 06 T = 2 P e 0 . 06 T = 2 ln e 0 . 06 T = ln 2 0 . 06 T = ln 2 Thus T = ln 2 / 0 . 06 11 . 55 An alternative way to solve this problem is to simply apply the formula for doubling time. Notice that in this case, k = 0 . 06 and Q 0 = A (0) = P . The doubling time is t d = ln 2 / 0 . 06 11 . 55 Therefore, we can conclude that it will take 12 years in order to double the value of the given investment.
30 J. S´ anchez-Ortega 2.9 New Functions from Old Functions In this section, we will learn how to obtain new functions by shifting, stretch- ing, and reflecting their graphs. As we will see, it will allow us to sketch the graphs of many functions quickly by hand. Translations Let c be a positive number . We can translate the graph of a function f horizontally and vertically as indicated in the table below.

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2. Functions 31 Example 2.19. The graph of y = x 2 + 1 is the parabola y = x 2 shifted upward 1 unit. The graph of y = x 2 - 3 is the parabola y = x 2 shifted downward 3 units. 0 1 -1 x y y = x 2 y = x 2 + 1 y = x 2 - 1 Example 2.20. The graph of y = ( x - 3) 2 is the parabola y = x 2 shifted 3 units to the right. The graph of y = ( x + 1) 2 is the parabola y = x 2 shifted 1 unit to the left. -1 3 0 x y y = x 2 y = ( x + 1) 2 y = ( x - 3) 2
32 J. S´ anchez-Ortega Example 2.21. Sketch the graph of y = x 2 - 4 x +3 shifting the graph of y = x 2 .

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