How do you find the critical points for $f(x)=3cos2(x-pi/4)+1$ ?

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Answer 1 · 2016-04-18T13:24:13

Use trigonometry to simplify the expression for $f(x)$ .

$cos2(x-pi/4) = cos(2x-pi/2) = sin2x$

(If you don't remember the identity above, use the difference formula for
$cos(2x-pi/2) = cos(2x)cos(pi/2)+sin(2x)sin(pi/2) = sin(2x)$ )

So $f(x) = 3sin(2x)+1$

and $f'(x) = 6cos(2x)$ which exists for all $x$ and is $0$ when

$cos(2x) = 0$ . Which is true exactly when

$2x = pi/2 + pik$ for integer $k$ . Or,

$x = pi/4 + pi/2k$ for integer $k$ .

Without rewriting we can get the same answer (of course).

$f(x) = 3cos2(x-pi/4)+1$

$f'(x) = -3sin2(x-pi/4)*[d/dx(2(x-pi/4)] = -3sin2(x-pi/4)*[2(1)]$

$f'(x) = -6sin2(x-pi/4)$

F'(x) is never undefined and it is $0$ for

$sin2(x-pi/4) = 0$ Which is true exactly when

$2(x-pi/4) = pik$ for integer $k$ . Or,

$x -pi/4= pi/2k$ for integer $k$ . So we need

$x = pi/4+ pi/2k$ for integer $k$ .

How do you find the critical points for f(x)=3cos2(x-pi/4)+1 f(x)=3cos2(x-pi/4)+1?