How do you find the critical numbers of $f(x)=sinxcosx$ ?

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Answer 1 · 2017-02-01T13:14:40

Critical numbers of $f(x)$ occur when

$x \ \ = pi/4 + (npi)/2 \ \ \ \ n in ZZ$
eg $x = +-pi/4, +-3pi/4, +-5pi/4, ...$

We have:

$f(x) = sinx cos x$

Differentiating wrt $x$ using the product rule:

$f'(x) = (sinx) (-sinx) + (cosx) (cosx)$
$" "= cos^2x -sin^2x$
$" "= cos(2x)$

At a critical point $f'(x)=0$

$:. cos (2x) = 0$
$:. 2x = pi/2 + npi$
$:. x \ \ = pi/4 + (npi)/2 \ \ \ \ n in ZZ$

Hence critical numbers of $f(x)$ occur when

$x \ \ = pi/4 + (npi)/2 \ \ \ \ n in ZZ$
eg $x = +-pi/4, +-3pi/4, +-5pi/4, ...$

We can see these values of $x$ correspond to max/min of the graph $y=sinx cos x$ :

enter image source here

How do you find the critical numbers of f(x)=sinxcosxf(x)=sinxcosx?