F(x)=sin(x+pi/2), 0<x= 5pi/4?

Question

f(x)=sin(x+pi/2), 0<x≤ 5pi/4
a. max is 1 at x=(7/8)pi, min is -1 at x=(1/8)pi
b. max is 1 at x=0,min is -1 at x=pi
c. max is 1 at x=(1/8)pi,min is -1 at x=(7/8)pi
d. max is 1 at x=(9/8)pi,min is -1 at x=(7/8)pi

Answer 1

The derivative F' is cos(x+pi/2) and that is zero when argument is pi/2 so that means x = 0 is a critical point.

 

But x=0 is NOT IN THE DOMAIN, as stated in the inequality given.

 

When x = pi, sin(3pi/2) = -1 and that looks like the minimum.

 

If the domain inequality had included 0, then I'd say the answer is b.  Otherwise, I say NONE OF THE ANSWERS a-d is correct.

Answer 2

Use the trig identity sin(A+B) = sinAcosB + cosAsinB:

 

f(x) = sin(x+π/2) = (sinx)(cos(π/2)) + (cosx)(sin(π/2))

 

                         = cosx

 

If 0 ≤ x ≤ 5π/4, cosx has a maximum value of 1 when x = 0 and a minimum value of -1 when x = π. 

Answer 3

Set the derivative of f(x) equal to zero.

 

f'(x) = 0

 

cos(x + π/2) = 0

 

Let    q = x + π/2

 

cos(q) = 0

 

q = π/2

q = 3π/2

 

Substitute  these values of q into the original substitution.

 

x + π/2 = π/2                and                x + π/2 = 3π/2

 

x = 0                           and                 x = π

 

 

These are your critical points.  Evaluate the derivative using the critical points.

 

If the derivative changes from positive to negative, then you have a maximum.

If the derivative changes from negative to positive, then you have a minimum.

 

Also, evaluate the f(0) and f(5π/4) to determine their values.