If $f : \mathbf{R} \to \mathbf{R}$ is a twice differentiable function such that $f''(x)>0$ for all $x \in \mathbf{R}$, and $f\left(\frac{1}{2}\right)=\frac{1}{2}, f(1)=1$, then

$f''(x) > 0 \ \forall x \in \mathbf{R} \ (\text{strictly convex})$

$f\!\left(\dfrac{1}{2}\right) = \dfrac{1}{2}, \qquad f(1) = 1$

$\text{Which is true about } f'(1)\,?$

$1.\ \text{draw a convex curve through the points}$

$2.\ \text{slope of the secant over } \left[\tfrac12, 1\right]$

$3.\ \text{use MVT and } f'' > 0$

$\text{secant slope} = \dfrac{f(1) - f(1/2)}{1 - 1/2} = \dfrac{1 - \tfrac12}{\tfrac12} = 1$

$\exists c \in \left(\tfrac12, 1\right): \ f'(c) = \dfrac{f(1)-f(1/2)}{1-1/2} = 1$

$f'' > 0 \ \Rightarrow\ f' \text{ strictly increasing}$

$c < 1 \ \Rightarrow\ f'(1) > f'(c) = 1 \ \Rightarrow\ f'(1) > 1$

$f'(1) > 1 \ \Rightarrow\ \text{option (4)}$