Suppose $f(x)$ is twice differentiable for all real $x$. Further suppose $|f(x)|\leqslant$ 1 and $|f''(x)| \leqslant$ 1 for all real $x$. Then $|f'(x)|$ is:
Options: a) $\leqslant$ 1$\quad$ b) $\gt$ 2 $\quad$ c) $\leqslant \frac 32$ $\quad$ d) $\gt \frac 32$
My thoughts: I first thought it could be $sin(x)$ or $cos(x)$ as it satisfied all the given conditions. I looked up the answer, and it was c. Since I had information about $f$ and $f''$ I tried applying Lagrange's Mean Value theorem on $f(x)$ and $f'(x)$ assuming arbitrary parameters to get two inequalities. $$\frac{f(b)-f(a)}{b-a} = f'(c)$$ for some arbitrary $a,b$ and $c$, and $c \; \epsilon \; (a,b)$. Similarly I applied the same theorem for $f'(x)$. Using the given conditions I could get inequalities using the given conditions on the values of $f(x)$ and $f''(x)$. I could not proceed further.
As I commented, the key is Taylor's theorem, not the MVT.
Solution.$\blacktriangleleft$. For an arbitrary $h >0$, consider Taylor's formula with Lagrange's remainder: \begin{align*} f(x + h) &= f(x) + h f'(x) + \frac {h^2}2 f''(c_1)\quad [c_1 \in (x, x+h)],\\ f(x - h) &= f(x) - hf'(x) + \frac {h^2}2 f''(c_2) \quad [c_2 \in (x-h, x)]. \end{align*} Subtracting them yields $$ f(x+h) - f(x-h) = 2h f'(x) + \frac {h^2}2 (f''(c_1) - f''(c_2)). $$ Now estimate $|f'(x)|$: \begin{align*} |f'(x)| &= \left|\frac {f(x+h) - f(x-h)} {2h} + \frac h 4 (f''(c_1) - f''(c_2))\right|\\ &\leqslant \frac {|f(x+h)| + |f(x-h)|} {2h} + \frac h 4 (|f''(c_1)| + |f''(c_2)|) \\ &\leqslant \frac 2{2h} + \frac h 4 \cdot 2\\ &= \frac 1h + \frac h2. \end{align*} Therefore, $$ \sup_{x\in \mathbb R}|f'(x)| \leqslant \frac 1h + \frac h2 \quad [\forall h > 0]. $$ By AM-GM inequality, the RHS has a minimum $2 \sqrt{(1/h) \cdot (h/2)} = \sqrt2$, thus $$ |f'(x)| \leqslant \sqrt 2 \doteq 1.414 \fbox{$\leqslant \frac 32$}, $$ hence the choice $(\mathrm C)$. $\blacktriangleright$