Find the global extrema of the function

Question

find the global extrema of the function f (x) = cos(2x) − x for x ∈ [− π/2 , π/2 ].

Find the critical points of f(x): f′(x)=−2sin(2x)−1

−2sin(2x)−1=0

2sin(2x)=−1

sin(2x)=−1/2

x=2x/2​=−7π/12​

Evaluate f(x) at the critical point and endpoints:​ f(−π/2​)=cos(2⋅(−π/2​))−(−π/2​)=1+π/2

f(π/​)=cos(2⋅π/2​)−π/2​=−1−/2π

f(x1​)=f(−7π/12​).​

​

f(−π​/2)=1+π/2​ is the maximum value, and f(π/2​)=−1−π/2​​ is the minimum value

am I doing this question correctly?

Answer 1

You are on the right track, but solving the trig equation has some issues.

sin(2x) = -1/2

2x = sin^-1(-1/2)

x = sin^-1(-1/2)/2

Angles that have a sine of -1/2 include -π/6 + 2kπ, and -5π/6 + 2kπ, k any integer. If you divide those angles by 2 to solve for x:

x = -5π/12 + kπ or -π/12 + kπ, k any integer.

If you let k take on values from -3 to 3 (for example), you will see that the only values ending between -π/2 and π/2 are when k = 0.

So to check for max and min determine the values of the function as follows:

f(-π/2)

f(-5π/12)

f(-π/12)

f(π/2)

Check here for confirmation:

desmos.com/calculator/n5owft5lxi