Calculus

Download PDF by Daniel A. Murray: A First Course in Infinitesimal Calculus

By Daniel A. Murray

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Example text

This yields ln(2x+1 ) = ln(51−2x ) (x + 1) ln 2 = (1 − 2x) ln 5 x ln 2 + ln 2 = ln 5 − 2x ln 5 x ln 2 + 2x ln 5 = ln 5 − ln 2 ln 5 − ln 2 ln 5 − ln 2 . x = ln 2 + 2 ln 5 x(ln 2 + 2 ln 5) = Using properties of logarithms, the answer can be simplified further as x == ln 52 ln 25 ln 52 ln 5 − ln 2 = = = . 2 2 ln 2 + 2 ln 5 ln 2 + ln 5 ln(2 · 5 ) ln 50 57. Change the logarithmic equation log2 (2x + 1) = 3 into the equivalent exponential equation 2x + 1 = 23 = 8. Then 2x = 7, or x = 7 2 . 58. Change the logarithmic equation log3 (3x − 2) = 2 into the equivalent exponential equation 3x − 2 = 32 = 9.

In the figure below, the graph of f is represented by the dashed curve, and the graph of P is represented by the solid curve. 5 -1 (d) The graph of H(x) = f (x + 1) − 2 can be obtained from the graph of f by shifting horizontally left 1 unit and vertically down 2 units . In the figure below, the graph of f is represented by the dashed curve, and the graph of H is represented by the solid curve. 1 -4 -3 -2 -1 1 2 3 -1 -2 -3 (e) The graph of Q(x) 1 2 f (x) factor of 12 = can be obtained from the graph of f by compressing vertically by a .

The domain of f −1 is the set {x|x ≥ 4} , and the range of f −1 is the set {y|y ≥ 0} . 38. (a) Start by writing y = f (x) = (x − 2)2 + 4, where x ≤ 2. Next interchange the x and y variables to obtain an implicit definition of the inverse function: x = (y − 2)2 + 4, where y ≤ 2. To obtain the inverse function explicitly, solve the implicit equation for y: √ (y − 2)2 = x − 4, so that y = 2 ± x − 4. √ Because y ≤ 2, only the negative square root is appropriate; thus, f −1 (x) = 2 − x − 4 . To check that this is the correct formula for the inverse function, show that f (f −1 (x)) = x and f −1 (f (x)) = x: √ √ f (f −1 (x)) = f (2 − x − 4) = (2 − x − 4 − 2)2 + 4 √ = (− x − 4)2 + 4 = x − 4 + 4 = x , f −1 (f (x)) = f −1 ((x − 2)2 + 4) = 2 − = 2− (x − 2)2 + 4 − 4 (x − 2)2 = 2 − |x − 2| = 2 − (2 − x) = x , where, in both calculations, attention is restricted to x ≤ 2.

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A First Course in Infinitesimal Calculus by Daniel A. Murray


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