2.5.16-eg-1 - Example Where is the function f (x) =...

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Unformatted text preview: Example Where is the function f (x) = continuous on the interval − x2 + 1 sin 2x , x∈ − 5π 5π , , 44 5π 5π ? , 44 Solution: The function f is continuous where its denominator is nonzero. We know that sin y = 0 at y = 0, ±kπ, k = 1, 2, . . . , so sin 2x = 0 at 2x = 0, ±kπ, k = 1, 2, . . . , that is, for x values satisfying π x = 0, ± k · , k = 1, 2, . . . . 2 5π 5π , , it follows that sin 2x = 0 at x = −π, −π/2, 0, π/2, π . So the 44 function f is continuous for So in the interval − x∈ − 5π 5π π π , , x = −π, − , 0, − , π, 44 2 2 or rewriting the above as a union of intervals, x∈ − 5π π π π 5π π , −π ∪ −π, − ∪ − , 0 ∪ 0, ∪ , π ∪ π, . 4 2 2 2 2 4 Note: In WeBWorK notation, the answer is: [-5pi/4,-pi) U (-pi,-pi/2) U (-pi/2,0) U (0,pi/2) U (pi/2,pi) U (pi, 5pi/4] ...
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