How do you find the critical points for $f(x)=sinx cosx$ and the local max and min in #0 ≤ x < 2pi?

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Answer 1 · 2017-01-24T14:39:25

$x = pi/4$ : Local max
$x = (3pi)/4$ : Local Min

Differentiate:

$f'(x) = cosx(cosx) + sinx(-sinx)$

$f'(x) = cos^2x - sin^2x$

$f'(x) = cos2x$

Critical points will occur when the derivative equals $0$ or is undefined.

There will be no undefined points on the function, since it's continuous.

$0 = cos2x$

$2x = pi/2 and (3pi)/2$

$x= pi/4 and (3pi)/4$

Now check the signs to the left and right of each of these points.

$f'(0) = cos(2(0)) = cos(0) = 1$

$f'(pi/2) = cos(2(pi/2)) = cos(pi) = -1$

So $x = pi/4$ is a local maximum (the derivative is increasing, then decreasing).

$f'((2pi)/3) = cos(2((2pi)/3)) = cos((4pi)/3) = -1/2$

$f'((5pi)/6) = cos(2((5pi)/6)) = cos((5pi)/3) = 1/2$

So, $x = (3pi)/4$ is a local minimum.

Hopefully this helps!

How do you find the critical points for f(x)=sinx cosx f(x)=sinx cosx and the local max and min in #0 ≤ x < 2pi?