If $f'(x)geq 1$ for all $xin[0,1]$ and $f$ is twice differentiable which of the following is true?

For every function f:[0,1] $tomathbb{R}$ which is twice differentiable and satisfies $f'(x)geq 1$ for all $xin[0,1]$, we must have

a)$f”(x)geq 0$ for all $xin[0,1]$

b) $f(x)geq x$ for all $xin[0,1]$

c) $f(x_2)-x_2leq f(x_1)-x_1$ for all $x_1,x_2in[0,1]$ with $x_2geq x_1$

d) $f(x_2)-x_2geq f(x_1)-x_1$ for all $x_1,x_2in[0,1]$ with $x_2geq x_1$

My thoughts: If we take $f(x)=ax+b$ and $f(x)=e^{ax}$ with $a>1$, option (a) is correct, option (b) won’t work for $b<0$, so (b) is not true. Option (d) is correct.

So, how do I confirm further between (a) and (d) ? What is a proper way to solve this (rather than case-by-case check) ?

Answer

Proof of (d)

Using Lagrange’s Mean Value Thm –

$frac{ f(x_2)- f(x_1)}{x_2 – x_1} = f'(c) $ where $x in (x_1,x_2) implies c in [0,1]$

Now, $f'(x) geq 1 forall x in [0,1]$

Thus $f'(c)geq 1$

And you get (d )after some trivial manipulations .

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