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anonymous

  • one year ago

What is the x-coordinate for the minimum point in the function f(x) = 4 cos(2x − π) from x = 0 to x = 2π? x = pi over 2 x = pi over 4 x = 2π x = π

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  1. anonymous
    • one year ago
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    @aloud @phi @mathmate

  2. phi
    • one year ago
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    do you know what cos looks like from 0 to 2pi? https://www.google.com/search?q=cos%28x%29&gbv=2&sei=FSSEVeKzLIevsAWqqoKYAQ

  3. anonymous
    • one year ago
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    i thought it intercepts at -4

  4. phi
    • one year ago
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    the smallest value will be y= -4 for a "normal cosine" from 0 to 2pi, the minimum value is at pi (exactly half way from 0 to 2pi)

  5. anonymous
    • one year ago
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    no I dont know how to graph them that well

  6. phi
    • one year ago
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    so you want to find where the "angle" of your problem f(x) = 4 cos(2x − π) from x = 0 to x = 2π equals pi in other words, what is "x" so that 2x - pi = pi

  7. anonymous
    • one year ago
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    would you plug in the x=0 for the x in 2x-pi=pi

  8. anonymous
    • one year ago
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    like substitution

  9. phi
    • one year ago
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    The easiest way is to look at a graph: https://www.google.com/search?q=cos%28x%29&gbv=2&sei=FSSEVeKzLIevsAWqqoKYAQ#q=4+cos%282x+%E2%88%92+%CF%80%29

  10. phi
    • one year ago
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    but using algebra 2x - pi = pi add +pi to both sides 2x - pi + pi = pi+pi on the left -pi+pi add up to zero. on the right you get 2pi 2x = 2pi now divide both sides by 2

  11. anonymous
    • one year ago
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    it would be x=pi because the 2's would cancel right?

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