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Review We first start with the properties of the graph of the basic logarithmic function of base a, f (x) = log a (x) , a > 0 and a not equal to 1. The domain of function f is the interval (0 , +inf). The range of f is the interval (-inf , +inf). The symbol inf means infinity. Function f has a vertical asymptote given by
x = 0. This function has an x intercept at (1 , 0). f increases as x increases.
Example 1: f is a function given by f (x) = log2 (x + 2)
Answer to Example 1 a - The domain of f is the set of all x values such that x + 2 > 0 or x > -2 The range of f is the interval (-inf , +inf). b - The vertical asymptote is obtained by solving x + 2 = 0 which gives x = -2 As x approaches -2 from the right (x > -2) , f(x) decreases without bound. How do we know this? Let us take some values: f(-1) = log2 (-1 + 2) = log2 (1) = 0 f(-1.5) = log2 (-1.5 + 2) = log2 (1/2) = -1 f(-1.99) = log2 (-1.99 + 2) = log2 (0.01) which is approximately equal to -6.64 f(-1.999999) = log2 (-1.999999 + 2) = log2 (0.000001) which is approximately equal to -19.93. c - To find the x intercept we need to solve the equation f(x) = 0 log2 (x + 2) = 0 Use properties of logarithmic and exponential functions to write the above equation as 2log2 (x + 2) = 20 Then simplify x + 2 = 1 x = -1 The x intercept is at (-1 , 0). The y intercept is given by (0 , f(0)) = (0,log2 (0 + 2)) = (0 , 1). d - So far we have the domain, range, x and y intercepts and the vertical asymptote. We need more points. Let us consider a point at x = -3/2 (half way between the x intercept and the vertical asymptote) and another point at x = 2. f(-3/2) = log2 (-3/2 + 2) = log2 (1/2) = log2 (2-1) = -1. f(2) = log2 (2 + 2) = log2 (22) = 2. We now have more information on how to graph f. The graph increases as x increases. Close to the vertical asymptote x = -2, the graph of f decreases without bound as x approaches -2 from the right. The graph never cuts the vertical asymptote. We now join the different points by a smooth curve.
Matched Problem to Example 1: f is a function given by f (x) = log2 (x + 3)
Example 2: f is a function given by f (x) = -3ln (x - 4)
Answer to Example 2 a - The domain of f is the set of all x values such that x - 4 > 0 or x > 4 The range of f is the interval (-inf , +inf). b - The vertical asymptote is obtained by solving x - 4 = 0 or x = 4 As x approaches 4 from the right (x > 4) , f(x) increases without bound. How do we know this? Let us take some values: f(5) = ln(5-4) = -3ln(1) = 0 f(4.001) = -3ln(0.001) which is approximately equal to 20.72. f(4.000001) = -3ln(0.000001) which is approximately equal to 41.45. c - To find the x intercept we need to solve the equation f(x) = 0 -3ln(x - 4) = 0 Divide both sides by -3 to obtain ln(x - 4) = 0 Use properties of logarithmic and exponential functions to write the above equation as eln(x - 4) = e0 Then simplify x - 4 = 1 x = 5 The x intercept is at (5 , 0). The y intercept is given by (0 , f(0)). f(0) is undefined since x = 0 is not a value in the domain of f. There is no y intercept. d - So far we have the domain, range, x intercept and the vertical asymptote. We need extra points to be able to graph f. f(4.5) = -3ln(4.5 - 4) approximately equal to 2.08 f(8) = -3ln(8 - 4) approximately equal to - 4.16 f(14) = -3ln(14 - 4) approximately equal to - 6.91 Let us now sketch all the points and the vertical asymptote. Join the points by a smooth curve and f increases as x approaches 4 from the right.
Matched Problem to Example 2: f is a function given by f (x) = 2ln (x + 5)
Example 3: f is a function given by f (x) = 2ln (| x |)
Answer to Example 3 a - The domain of f is the set of all x values such that | x | > 0 The domain is the set of all real numbers except 0. The range of f is the interval (-inf , +inf). b - The vertical asymptote is obtained by solving | x | = 0 which gives x = 0 As x approaches 0 from the right (x > 0) , f(x) decreases without bound. How do we know this? Let us take some values: f(1) =2 ln(| 1 |) = 0 f(0.1) = 2ln(0.1) which is approximately equal to -4.61. f(0.0001) = 2ln(0.0001) which is approximately equal to -18.42. f(0.0000001) = 2ln(0.0000001) which is approximately equal to -32.24. As x approaches 0 from the left (x < 0) , f(x) decreases without bound. How do we know this? Let us take some values: f(-1) =2 ln(| -1 |) = 0 f(-0.1) = 2ln(| -0.1 |) which is approximately equal to -4.61. f(-0.0001) = 2ln(| -0.0001 |) which is approximately equal to -18.42. f(-0.0000001) = 2ln(| -0.0000001 |) which is approximately equal to -32.24. c - To find the x intercept we need to solve the equation f(x) = 0 2ln(| x |) = 0 Divide both sides by 2 to obtain ln(| x |) = 0 Use properties of logarithmic and exponential functions to write the above equation as eln(| x |) = e0 Then simplify | x | = 1 Two x intercepts at (1 , 0) and (-1 , 0). The y intercept is given by (0 , f(0)). f(0) is undefined since x = 0 is not a value in the domain of f. There is no y intercept. d - So far we have the domain, range, x intercept and the vertical asymptote. By examining function f it is easy to show that this is an even function and its graph is symmetric with respect to the y axis. f( -x) = 2 ln(| -x |) but | -x | = | x | and therefore f( -x) = 2 ln(| x |) = f( x ), this shows that f is an even function. Let us find extra points. f(4) = 2ln(| 4 |) approximately equal to 2.77. f(0.5) = 2ln(| 0.5 |) approximately equal to - 1.39. Since f is even f(-4) = f(4) and f(-0.5) = f(0.5). Let us now sketch all the points, the vertical asymptote and Join the points by a smooth curve.
Matched Problem to Example 3: f is a function given by f (x) = -2ln (x2)
More references on logarithmic functions and graphing. |
