Show that $f(x) \geq 1 -x$ if $f(0) \geq 1 $ and $f'(x) \geq -1$

Question

Let $f:[0, \pi] \rightarrow \mathbb{R}$ continuous and differentiable function in $[0, \pi]$ with $f(0) \geq 1 $ and $f'(x) \geq -1$ for all $x \in ]0, \pi]$. Show that $f(x) \geq 1 -x$ for all $x \in [0, \pi]$ with mean value theorem. Is this function monoton ?

So from the mean value theorem follows that a x exists such that $f'(x) =\frac{f(\pi) - f(0) }{ \pi - 0 }$. So $\frac{f(\pi) - f(0)}{ \pi} \geq -1$ $\Rightarrow $ $f(\pi) \geq -\pi + f(0)$ since $f(0) \geq 1$ it follows that $f(\pi) \geq -2.141....$ But now I don't know what to do.

Answer 1

Correct me if wrong:

MVT:

$\dfrac{ f(x)-f(0)}{x-0} =f'(t)$, $t \in (0,x),$

$f(x) = f'(t)x +f(0) \ge$

$ f'(t)x +1 \ge 1-x.$

Used:

1)$f(0) \ge 1$; and

2) $f'(t)x \ge -x$, since

$f'(t) \ge -1$, and $x \ge 0$.

Monotonic?

Example: $y = (x-1/2)^2 +3/4.$

$y(0)=1$; and

$y'(x)= 2(x-1/2) =$

$ 2x - 1 \ge -1$ for $x \in [0,π].$

Note : The vertex is at : $x = 1/2.$

Answer 2

\begin{align*} f(x)&=f(x)-f(0)+f(0)\\ &=f'(\xi)x+f(0)\\ &\geq f'(\xi)x+1\\ &\geq -x+1. \end{align*}

Answer 3

We have for $x\in [0, \pi]$

$$\frac{f(x) - f(0) }{ x - 0 }=f'(c)\implies f(x)=xf'(c)+f(0)\ge1-x$$

from $f'(x)\ge -1$ we can't conclude that the function is monotone.