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