# 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 .