So really the question is find the minimum value of f(x) = 12 / (cosx + √3 sinx + 4)
and there's no degree sign, so I'm assuming the answer should be in radians, though use degrees when explaining, cause I'm not good with radians (blasted things).
Anyhow, I'm sure if you can tell me what the minimum value of "cosx + √3 sinx" is, and why the minimum value is what it is, then I'd be able to work out the rest.
and there's no degree sign, so I'm assuming the answer should be in radians, though use degrees when explaining, cause I'm not good with radians (blasted things).
Anyhow, I'm sure if you can tell me what the minimum value of "cosx + √3 sinx" is, and why the minimum value is what it is, then I'd be able to work out the rest.
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Actually, f(x) = 12/[cos(x) + √3*sin(x) + 4] is minimized when cos(x) + √3*sin(x) + 4 is maximized, so you actually want the maximum value of cos(x) + √3*sin(x) + 4 and not its minimum value. There are two ways to do this: with and without Calculus. The non-Calculus approach is to write:
cos(x) + √3*sin(x) = a*sin(x + b), for some a and b.
Since -a ≤ cos(x) + √3*sin(x) ≤ a, it suffices to only find the value of a.
By expanding a*sin(x + b):
a*sin(x + b)
= a[sin(x)cos(b) + cos(x)sin(b)], by the sine addition formula
= [a*cos(b)]sin(x) + [a*sin(b)]cos(x).
By comparing coefficients, we have:
a*cos(b) = √3 and a*sin(b) = 1.
By squaring both sides of both equations and adding:
[a*cos(b)]^2 + [a*sin(b)]^2 = (√3)^2 + 1^2
==> a^2[cos^2(b) + sin^2(b)]^2 = 4
==> a^2 = 4, since cos^2(b) + sin^2(b) = 1
==> a = 2.
Thus:
-2 ≤ cos(x) + √3*sin(x) ≤ 2 ==> 2 ≤ cos(x) + √3*sin(x) + 4 ≤ 6,
and:
12/[cos(x) + √3*sin(x) + 4] ≥ 12/6 = 2.
The Calculus approach is to let:
g(x) = cos(x) + √3*sin(x).
Then, by differentiating:
g'(x) = √3*cos(x) - sin(x).
Then, this equals zero when:
√3*cos(x) - sin(x) = 0
==> sin(x) = √3*cos(x)
==> tan(x) = √3, by dividing both sides by cosine
==> x = π/3 ± πk, where k is an integer.
With g''(x) = -√3*sin(x) - cos(x), we see that:
(a) g''(π/3) = -√3*sin(2π/3) - cos(2π/3) = -2 < 0
(b) g''(4π/3) = -√3*sin(4π/3) - cos(4π/3) = 2 > 0.
(Note that we do not need to test any for k > 1 because sine and cosine are periodic.)
Thus, the maximum value of g(x) = cos(x) + √3*sin(x) + 4 is:
g(π/3) = cos(π/3) + √3*sin(π/3) + 4 = 6,
and f(x)'s minimum value is 12/6 = 2.
I hope this helps!
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It's either that you want to know the maximum value or the answer key is wrong. Take a look at the graph here:
http://www.wolframalpha.com/input/?i=12%…
cos(x) + √3*sin(x) = a*sin(x + b), for some a and b.
Since -a ≤ cos(x) + √3*sin(x) ≤ a, it suffices to only find the value of a.
By expanding a*sin(x + b):
a*sin(x + b)
= a[sin(x)cos(b) + cos(x)sin(b)], by the sine addition formula
= [a*cos(b)]sin(x) + [a*sin(b)]cos(x).
By comparing coefficients, we have:
a*cos(b) = √3 and a*sin(b) = 1.
By squaring both sides of both equations and adding:
[a*cos(b)]^2 + [a*sin(b)]^2 = (√3)^2 + 1^2
==> a^2[cos^2(b) + sin^2(b)]^2 = 4
==> a^2 = 4, since cos^2(b) + sin^2(b) = 1
==> a = 2.
Thus:
-2 ≤ cos(x) + √3*sin(x) ≤ 2 ==> 2 ≤ cos(x) + √3*sin(x) + 4 ≤ 6,
and:
12/[cos(x) + √3*sin(x) + 4] ≥ 12/6 = 2.
The Calculus approach is to let:
g(x) = cos(x) + √3*sin(x).
Then, by differentiating:
g'(x) = √3*cos(x) - sin(x).
Then, this equals zero when:
√3*cos(x) - sin(x) = 0
==> sin(x) = √3*cos(x)
==> tan(x) = √3, by dividing both sides by cosine
==> x = π/3 ± πk, where k is an integer.
With g''(x) = -√3*sin(x) - cos(x), we see that:
(a) g''(π/3) = -√3*sin(2π/3) - cos(2π/3) = -2 < 0
(b) g''(4π/3) = -√3*sin(4π/3) - cos(4π/3) = 2 > 0.
(Note that we do not need to test any for k > 1 because sine and cosine are periodic.)
Thus, the maximum value of g(x) = cos(x) + √3*sin(x) + 4 is:
g(π/3) = cos(π/3) + √3*sin(π/3) + 4 = 6,
and f(x)'s minimum value is 12/6 = 2.
I hope this helps!
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It's either that you want to know the maximum value or the answer key is wrong. Take a look at the graph here:
http://www.wolframalpha.com/input/?i=12%…