+56 What are the maximum value and minimum values of f(x) = |2sin(2x - pi/3) - 5| + 3
+14 If you'd have to calculate it you would wanna calculate the derivative and see where that equals 0.
(To calculate the derivative of an absolute value function you have to split it into two for when 2sin(2x - pi/3) - 5 > 0 and 2sin(2x - pi/3) - 5 < 0 (and add a minus before this one))
if you simply need those values, just plot it and you'll get minimum 6, maximum 10
+1491 Are they approaching infinity? Or are they bound by an "a" or "b"
max{abs(2 sin(2 x - π/3) - 5) + 3} = 10 at x = -π/12 + n π for integer n
min{abs(2 sin(2 x - π/3) - 5) + 3} = 6 at x = -(7 π)/12 + n π for integer n
min{abs(2 sin(2 x - π/3) - 5) + 3} = 6 at x = (5 π)/12 + n π for integer n
+129850 This will have a max value where the sin = 1..this happens at pi/2.....this means that we need to find where [2x - pi/3] = pi/2
So......solving for x, we have that
2x - pi/3 = pi/2 add pi/3 to both sides
2x = [ 5pi/6] divide both sides by 2
x = 5pi / 12 .......and it will reach that max at every ± pi rads from this value
To find out where it has a min....we need to find out where the sin = -1...this happens at 3pi/2....thus, we need to find out where
2x - pi/3 = 3pi/2 add pi/3 to both sides
2x = 11pi/6 divide both sides by 2
x = 11pi/12 ........and it wil,. again, have a min at every ±pi rads from this value
See the graph, here : https://www.desmos.com/calculator/ohrm0oeuul
