What is the extreme f(x)=sin(x)cos(x)+2 ?

The extreme f(x)=sin(x)cos(x)+2 is Maximum(pi/4+pin, 5/2),Minimum((3pi)/4+pin, 3/2)

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extreme f(x)=sin(x)cos(x)+2

extreme points

f (x) = sin (x) cos (x) + 2

Maximum (\frac{π}{4} + πn, \frac{5}{2}), Minimum (\frac{3 π}{4} + πn, \frac{3}{2})

Solution steps

Find the critical points:

x = \frac{π}{4} + πn, x = \frac{3 π}{4} + πn

Domain of

sin (x) cos (x) + 2 : - \infty < x < \infty

Find intervals:

Increasing

: πn \leq x < \frac{π}{4} + πn,

Decreasing

: \frac{π}{4} + πn < x < \frac{3 π}{4} + πn,

Increasing

: \frac{3 π}{4} + πn < x < π + πn

Plug the extreme point

x = \frac{π}{4} + πn

into

sin (x) cos (x) + 2 \Rightarrow y = \frac{5}{2}

Maximum (\frac{π}{4} + πn, \frac{5}{2})

Plug the extreme point

x = \frac{3 π}{4} + πn

into

sin (x) cos (x) + 2 \Rightarrow y = \frac{3}{2}

Minimum (\frac{3 π}{4} + πn, \frac{3}{2})

Maximum (\frac{π}{4} + πn, \frac{5}{2}), Minimum (\frac{3 π}{4} + πn, \frac{3}{2})

f (x) = \frac{( x - 2 ) ^{2}}{25} + \frac{( y - 7 ) ^{2}}{36}

f (x) = (\frac{( x ^{2} - 1 ) ^{0.5}}{x ^{3}})

f (x) = x^{2} + x - 3 x y + y^{3} - 5

f (x, y) = 6 x^{2} + 9 x y + 3 y^{2} + 8 x y^{2} + 5 x^{2} y

\frac{x ^{2} - 11}{x - 6}

What is the extreme f(x)=sin(x)cos(x)+2 ?
The extreme f(x)=sin(x)cos(x)+2 is Maximum(pi/4+pin, 5/2),Minimum((3pi)/4+pin, 3/2)