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Suppose that $f$ is differentiable on $\mathbb{R}$. If $f(0)=1$ and |f^{'}(x)|\leq1 for all $x\in\mathbb{R}$, prove that $|f(x)|\leq|x|+1$ for all $x\in\mathbb{R}$.

I tried:

Let $g(x)=|f(x)|-|x|-1$. Then I tried to find g^{'}(x) but I'm not sure where to start.

2 Answers 2

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The Mean Value Theorem is what you're looking for.

\phantom{stuff to make this a nontrivial answer - way to go MSE. BTW, I'm hoping for a Patriots Win}

  • 0
    Now that a complete answer is given, I'll expand. Suppose, for instance, that f(2) > 3. For ease, suppose it's 4. Then there is a point in $(0,2)$ that has derivative \frac{4 - 1}{2 - 0} > 1. Contradiction.2012-01-25
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Integrate?

|f'(x)| \leq 1 implies that -1 \leq f'(x) \leq 1 integrate from 0 to $x$ and get $-x \leq f(x)-f(0) \leq x$ thus $-x+1 \leq f(x) \leq x+1$. If $f(x) \geq 0$, then $|f(x)|=f(x) \leq x+1 \leq |x|+1$. If $f(x)<0$, then $-x+1 \leq f(x)<0$ so $0<|f(x)|=-f(x)\leq x-1\leq|x-1|\leq |x|+1$. So either way, $|f(x)| \leq |x|+1$.