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Math HL range-domain based question


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Can someone explain me how to do part b and c

b) x = x, y = gif.latex?\frac{\pi}{6}. Use formula derived in a) and gif.latex?\cos{(2x)} = 1 - 2\sin^2{x} to solve and graph gif.latex?\sin^2{x}. Rewrite f(x) as a recognizable transformation of cos (2x).

Answer: [-1/4, 3/4]

c) Recognize that g(x) is the reciprocal function of f(x). That is g(x) = 1/f(x)

1/(3/4) = 4/3 = new bound

1/(-1/4) = -4 = new bound

But instead of g(x) lies between these bounds, g(x) lies outside of these bounds. (as is sin(x) and csc (x))

Answer: gif.latex?\left]-\infty,-4\right] \cup \

Edited by kw0573
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