find the maxima and minima of f(x) =sin x cos x +4

Question

applying the first derivative Test to identify all relative extrema

relative maxima (x,y) smallest value

                       (x,y) largest value

relative minima (x,y) smallest value

                       (x,y) largest value

Answer 1

f(x)=sin(x)cos(x)+4=½sin(2x)+4,

f'(x)=cos(2x)=0 at extrema, which occur when 2x=(2n+1)π/2 (n is an integer).

Therefore x=(2n+1)π/4.

f((2n+1)π/4)=½sin((2n+1)π/2)+4.

When n=0, 2, 4, ... f(x)=9/2; when n=1, 3, 5, ... f(x)=7/2.

f"(x)=-2sin(2x)<0 (maximum) when n=0, 2, 4, ...; and -2sin(2x)>0 (minimum) when n=1, 3, 5, ...

We can redefine the solution x by replacing n with 2n for 0, 2, 4, ... and with 2n+1 for 1, 3, 5, ...

x=(4n+1)π/4, f(x)=9/2 at maxima; x=(2(2n+1)+1)π/4=(4n+3)π/4 at minima. This gives us the maxima at (((4n+1)π/4,9/2) and minima at ((4n+3)π/4,7/2).