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Finding minimum value of x on an interval?
Within the interval (-2.63, 3.65), for what value of x does f(x) take on a minimum if
f(x)= −3.4*sin(x) + 1.9*cos(x)?
I know to take the derivative, set it equal to zero, and solve for x - I divided both sides by cos(x) to get a tan(x) and to cancel the other cos(x).
Now I have tan(x) = -3.4/1.9, or -34/19. I can take the inverse tangent, but there are an infinite number of answers to that function. That's where the interval comes in, but I'm not sure how many times I need to add 2*pi for the period, because it said it was wrong when I did that before.
Anyone see what I did wrong?
2 Answers
- ted sLv 72 years ago
you could also write f(x) = A sin ( x + Θ)....A² = 3.4² + 1.9²....tan Θ = 1.9 / [ - 3.4]...then you know this is min when x + Θ = π......{ - 1.06 & 2.08 are the critical numbers }...and period of tan is π
- mizooLv 72 years ago
f(x)= −3.4*sin(x) + 1.9*cos(x)
f'(x) = -3.4 cos x - 1.9 sin x
Wee need to find f'(x) = 0 and f(-2.63) and f(3.65)
f(-2.63) = f(3.65) ≈ 0
f'(x) = 0
-3.4 cos x - 1.9 sin x = 0
tan x = -3.4/1.9
x = Arctan -3.4/1.9
x ≈ -1.06
General solutions: x = 3.14n -1.06, n any integer
Solve the equation for different n's and see which one(s) are in the given interval.
(-2.63, 3.65)
n = -1, x = -4.02
n = 0, x = -1.06
n = 2, x = 2.08
n = 3, 8.36
=> x = -1.06, 2.08
f(-1.06) ≈ 3.89 => the maximum value
f(2.08) ≈ -3.89 => the minimum value
