Answer
See explanation.
Work Step by Step
(a) $f^{\prime}(x)=2 \cos x \sin x=\sin 2 x$ , $f^{\prime \prime}(x)=2 \cos 2 x$
$f^{\prime}(x)<0$ if $x<0$ and $f^{\prime}(x)>0$ if $x>0,$ so according to the first derivative test $0=x$ is a relative minimum.
$f^{\prime}(0)=0$ and $f^{\prime \prime}(0)=2>0,$ according to the second derivative test $0=x$ is a relative minimum.
(b) $f^{\prime}(x)=2\sec ^{2} x \tan x $ and $f^{\prime \prime}(x)=-4 \tan x \sec x \sec x \tan x+2 \sec ^{4} x=$
$2 \sec ^{2} x\left(\sec ^{2} x-2 \tan ^{2} x\right)$
$f^{\prime}(x)<0$ if $x<0$ and $f^{\prime}(x)>0$ if $x>0,$ so according to the first derivative test, $0=x$ is a relative minimum.
$f^{\prime}(x)=0$ if $x<0$ and $f^{\prime \prime}(0)>0,$ so according to the first derivative test, $0=x$ is a relative minimum.
(c) The function is non-negative. Since the functions are zero at $x=0$ and the functions are non-negative, we know that the functions need to have a relative minima at $0=x$
