Calculus 1c-2, Examples of Elementary Functions by Mejlbro L.

By Mejlbro L.

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I. It follows directly from cos 2y = 1 − 2 sin2 y that cos(2 Arcsin x) == 1 − 2 sin2 (Arcsin x) = 1 − 2x2 , x ∈ [−1, 1]. 6 Find the complete solution x = ϕ(t) of the differential equation dx 1 + x2 = . dt 1 + t2 Hint: Use that tan(u + v) = tan u + tan v . 1 − tan u · tan v A. A nonlinear differential equation of first order, where the variables can be separated. D. Solve the differential equation by separating the variables. I. e. Arctan x = Arctan t + k. We shall now use the hint with c = tan k, 10 8 6 y 4 2 –3 –2 0 –1 2 1 3 x –2 –4 –6 –8 –10 Figure 23: Some solution curves of 1 + x2 dx = .

I. 1) If x = f (y) = tan y, y ∈ − , , then 2 2 dx = 1 + tan2 y > 0, dy and we conclude that the inverse function exists, y = g(x) = Arctan x, x ∈ R. We get according to the theorem of differentiating an inverse function, g (x) = 1 1 1 = . 5 0 1 2 3 4 x Figure 30: The graph of y = x− Arctan x, x > 0. 2) Let f (x) = x− Arctan x, x ∈ R. Then, according to (1), f (x) = 1 − 1 x2 = >0 1 + x2 1 + x2 for x = 0. com 61 Calculus Analyse c1- 2 The Arcus Functions Since f (0) = 0, we get 0 < f (x) = x − Arctan x for x > 0, (and 0 > f (x) = x− Arctan x for x < 0), hence by a rearrangement, Arctan x < x for x > 0.

Com 43 Calculus Analyse c1- 2 Inverse functions, general b) If instead a ∈ R+ \ {1}, we get by taking the logarithm, ln y = x · ln a. Since ln a = 0 for a ∈ R+ \ {1}, the inverse function is x= ln y , ln a x ∈ R, a ∈ R+ \ {1}. 3 Argue why the function y = x3 + x2 + x + 1, x ∈ R, has an inverse function x = ϕ(y), y ∈ R. Find ϕ (4). A. Inverse function. D. e. ψ (x) > 0. 5 1 x –1 Figure 22: Part of the graph of y = x3 + x2 + x + 1. I. The function ψ(x) = x3 + x2 + x + 1, x ∈ R is of class C ∞ , and ψ (x) = 3x2 + 2x + 1 = 3 x2 + 1 2 x+ 3 9 +1− 1 1 =3 x+ 3 3 2 + 2 > 0.

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