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**Extra info for Calculus 1c-2, Examples of Elementary Functions**

**Example text**

Find the domains of two formulæ. D. Check the deﬁnition, the domain and the range for the given functions. I. We know that Arccos is deﬁned by y = cos x ⇔ when x ∈ [0, π] and y ∈ [−1, 1]. x = Arccos y, 1) Let x ∈ [0, π], and put y = cos x ∈ [−1, 1]. Then x = Arccos y = Arccos(cos x), x ∈ [0, π]. Since Arccos(cos x) ∈ [0, π], we conclude that the formula is never right, when x ∈ / [0, π]. Hence, the formula is correct, if and only if x ∈ [0, π]. 2) The left hand side is only deﬁned when x ∈ [−1, 1].

C = −π for x ∈ ] − ∞, −1[, and Arcsin 2x 1 + x2 = −π − 2 Arctan x, x ∈ ] − ∞, −1[. All things put together we see that ⎧ ⎨ π − 2 Arctan x, x ∈ ]1, +∞[, 2x 2 Arctan x, x ∈ ] − 1, 1[, Arcsin = ⎩ 1 + x2 −π − 2 Arctan x, x ∈ ] − ∞, −1[. Now, f (x) is continuous for every x ∈ R, and all the right hand sides are continuous in each their domains. Then the formulæ must by continuity also be valid at the end points. Then by a rearrangement, ⎧ 2x ⎪ ⎪ , x ∈ [1, +∞[, ⎪ π − Arcsin ⎪ ⎪ 1 + x2 ⎪ ⎨ 2x (2) 2g(x) = 2 Arctan x = , x ∈ [−1, 1], Arcsin ⎪ 1 + x2 ⎪ ⎪ ⎪ 2x ⎪ ⎪ , x ∈ ] − ∞, −1].

GCHQ values diversity and welcomes applicants from all sections of the community. We want our workforce to reflect the diversity of our work. com 56 Calculus Analyse c1- 2 The Arcus Functions 2) Similarly we get 1 · Arctan x dx = x · Arctan x − = x · Arctan x − 1 dx 1 + x2 x· 1 ln 1 + x2 . 2 C. Test. We get by a diﬀerentiation d 1 x · Arctan x − ln 1 + x2 dx 2 2x 1 x − · = Arctan x. 11 Find the maximal domains of the functions (1) f (x) = Arccos(cos x), (2) cos(Arccos x), and sketch their graphs.