By Alfred North Whitehead

Whitehead explains in extensive phrases what arithmetic is set, what it does, and the way mathematicians do it.Generations of readers who've stayed with the thinker from the start to the top have came upon themselves amply rewarded for taking this trip. As

*The big apple Times*saw many years in the past, "Whitehead does not popularize or make palatable; he's easily lucid and cogent ... A finely balanced mix of wisdom and urbanity .... may still pride you."

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Thus f ðxÞ ! À1. These can also be combined into the system: 8 gðxÞ À1 > < f ðxÞ ! À1 > : gð x Þ ¼ f ð x Þ This system has a solution iff (if and only if) x ¼ À3=4 ¼ À0:75. gðxÞ ¼ f ðxÞ ¼ À1 and Answer x ¼ À0:75. Problem 27 Does the equation 2 sin solutions? 2 x þ 2 cos 2 x ¼ 1:5ð tan x þ cot xÞ have any Solution Let us rewrite the equation in the form f ðxÞ ¼ gðxÞ. Applying pﬃﬃﬃﬃﬃ the formula a þ b ! 2 ab to the left and right sides of the equation and replacing cos 2 x ¼ 1 À sin 2 x in the exponent, we obtain the lower bounds for both sides of the equation f ðxÞ ¼ 2 sin 2 x þ 2 cos 2 x ¼ 2 sin 2 x gðxÞ ¼ 1:5ð tan x þ cot xÞ !

Let us practice solving problems. 8 f ðxÞ ¼ 2þ3x 2 4 3 y 2 1 -10 -5 0 5 x -1 10 14 1 Solving Problems Using Properties of Functions Problem 7 Is the function f ðxÞ ¼ function? pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 4 À jxj a bounded or unbounded Solution Let b be a possible value of the function f(x). Let us ﬁnd the solution to pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ the equation 4 À jxj ¼ b where b ! 0. Squaring both sides, we obtain 4 À jxj ¼ b2 or jxj ¼ 4 À b2 . Since jxj ! , 4 À b2 ! 0. 0 Therefore the function is bounded, 0 f ðx Þ b 2: 2.

If both cosine terms equal one, the function will obtain a maximum at that x value. To determine if this occurs, we attempt to solve the system: 8 ( < x ¼ 2πn, n 2 Z cos x ¼ 1 ) x ¼ 2πn, n 2 Z: , 2πm :x ¼ ,m 2 Z cos 3x ¼ 1 3 So the distance between two neighboring maxima is 2π. Answer The minimal period is 2π. Problem 14 Let f(x) be a periodic function with period T ¼ 13. Evaluate f(1) if À10Á 2 f 2 ð2Þ À 5 f ð0Þ þ 21 4 ¼ 0 and 4 f ðÀ1Þ À 4 f 3 ¼ 35. Solution This is an unusual problem and in order to solve it we need to apply the properties of periodic functions.