Set of slopes of tangents and secants of a continuous function Given a continuous function $f$ let us define the following subsets of the set $R$ of real numbers.
$T$ =set of slopes of all possible tangents to the graph of $f$. 
$S$ =set of slopes of all possible secants, i.e. the lines joining two points on the graph of $f$.
The question is to examine whether the following statements are true or false.
(i) If $f$ is differentiable, then $S$ is a subset of $T$. 
(ii) If $f$ is differentiable, then $T$ is a subset of $S$. 
(iii) If $T = S = R$, then $f$ must be differentiable everywhere. 
(iv) Suppose $0$ and $1$ are in $S$. Then every number between $0$ and $1$ must also be in $S$.
I have no idea on how to even approach this problem.The official solution hints for the use of mean value theorem and I have no idea on how it should be applied here.Any idea shall be highly appreciated.Thanks.
 A: I'll assume $f: \mathbb R \to \mathbb R$ is continuous, and that "differentiable" means differentiable at each point of $\mathbb R.$
(i) As mentioned by several others, the statement here is true by the MVT.
(ii) The statement is false: For the function $f(x)=x^3,$ $T=[0,\infty)$ while $S= (0,\infty).$ (The reason $0\notin S$ is because $f$ is strictly increasing, implying all slopes of secants are positive.)
(iii) False: Define
$$f(x)=\begin{cases} \sqrt {1-x^2},& -1\le x \le 1\\\,\,\,\, 0, & |x|>1\end{cases}.$$
The graph of $f$ is the upper half of the unit circle together with the rays $(-\infty,-1]\times \{0\}, [1,\infty)\times \{0\}.$ Thus $f$ is continuous on $\mathbb R.$
Because $f'(x) = -x/\sqrt {1-x^2}$ for $x\in (-1,1),$ we have $f'((-1,1))=\mathbb R.$ Thus $T=\mathbb R.$ Considering slopes of secants through $(-1,0)$ and $(x,\sqrt {1-x^2})$ for $x\in (-1,1],$ we see $S$ contains $[0,\infty).$ Similarly, the secants through $(1,0)$ and $(x,\sqrt {1-x^2})$ for $x\in [-1,1)$ show $S$ contains $(-\infty,0].$ So we have $S = T = \mathbb R,$ while $f$ is not differentiable at $-1$ or $1.$
(iv) True: Let $U=\{(x,y)\in \mathbb R^2: y>x\}.$ Then $U$ is connected (in fact it's convex). Define $F:U\to \mathbb R$ by
$$F(x,y) = \frac{f(y)-f(x)}{y-x}.$$
Then $F$ is continuous on $U$ by the continuity of $f.$ Note that $S = F(U).$ Since $U$ is connected, $F(U)=S$ is connected by continuity. Hence $S$ is an interval. Since we are given $0,1\in S$ we must have $[0,1]\subset S.$
A: As already observed in the comments, the first statement is true: according to the mean value theorem, we know that  $ S \subseteq T \,$. Remind that, in its most intuitive formulation, the mean value theorem states that, if a function is continuous on a closed interval $[a,b] \,$ and differentiable on the open interval $(a,b) \,$, there exists a point $c $ in $(a,b) \,$ such that the tangent in $c $ is parallel to the secant drawn through the endpoints of the interval. 
The second statement is false: while $S$ is a subset of $T $, the converse is not true. For example, consider  $f (x)=\cos (x) \,\,\, $. The tangent lines in the points where the function crosses the $x $-axis have no parallel secants. So we have that the values $f'(x) =\pm 1 \,\,\,$ in these points are included in $T $ but not in $S $.
The third statement is false. We can have $ T = S =R\,\,\, \,$ for functions that are not differentiable everywhere.  Consider, for example, $f (x)=\sec (x) \,\,\,$.
The fourth statement is true. This can be considered a result of the intermediate value property of our continuous function $f $.
