Answer
(a)
(b)
Since the derivative of the function is nonnegative.
The slope of the tangent line will also be nonnegative.
In the graph, we can see that the slope is nonnegative.
(c)
$t=0,2\pi,4\pi$ in the interval $0\leq t \leq 4\pi$.
Work Step by Step
(a)
(b)
Firstly derivate $g(t)=t-\sin{t}$.
We get, $g'(t)=1-\cos{t}$
Since, $\cos{t}\leq1$
$0\leq1-\cos{t}$
$\implies 0\leq g'(t)$
Since the derivative of the function is nonnegative.
The slope of the tangent line will also be nonnegative.
In the graph, we can see that the slope is nonnegative.
(c)
If the tangent line is horizontal, then, the slope will be zero.
And if the slope is zero, then, the derivative at that point will also be zero.
That is, $g'(t)=0$
Now substitute $g'(t)=0$ in $g'(t)=1-\cos(t)$.
We get, $0=1-\cos{t}$
$\implies \cos{t}=1$
So, $t=0,2\pi,4\pi$ in the interval $0\leq t \leq 4\pi$.
