Answer
a) Horizontal tangents at $+ \frac{\pi}{2}$,$- \frac{\pi}{2}$ and $+ \frac{3\pi}{2}$,$- \frac{3\pi}{2}$.
b) Horizontal tangents at $+ \frac{\pi}{2}$ and $- \frac{3\pi}{2}$.
c) No tangent line.
d) Horizontal tangents at $+2\pi,-2\pi,+\pi,-\pi$ and $0$.
Work Step by Step
a) $f(x) = sinx$
Solution :
Taking derivative
$f'(x) = cosx$
putting $f'(x) = 0$
$cosx = 0$
$x = cos^{-1}(0)$
Horizontal tangents at $+ \frac{\pi}{2}$,$- \frac{\pi}{2}$ and $+ \frac{3\pi}{2}$,$- \frac{3\pi}{2}$.
b) $f(x) = x+ cosx$
Solution :
Taking derivative
$f'(x) = 1 - sinx$
putting $f'(x) = 0$
$1 - sinx = 0$
$1 = sinx$
$x = sin^{-1}(1)$
Horizontal tangents at $+ \frac{\pi}{2}$ and $- \frac{3\pi}{2}$.
c) $f(x) = tanx$
Solution :
Taking derivative
$f'(x) = sec^{2}x$
As $sec^{2}x \geq 1$ always
So , no tangent line.
d) $f(x) = secx$
Solution:
Taking derivative
$f'(x) = secxtanx$
$f'(x) = \frac{sinx}{cos^{2}x}$ = 0
When $sinx = 0$
Horizontal tangents at $+2\pi,-2\pi,+\pi,-\pi$ and $0$.
