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I was doing a question

$f:\mathbb{R} \rightarrow \mathbb{R}$ is such that $f(0) = 0$ and $|\frac{df}{dx}(x)| \leq 5$ $\forall$ x .We can conclude that $f(1)$ is in a)$(5,6)$ b) $[-5,5]$ c) $(-\infty,-5) \cup (5 ,\infty)$ d)$[-4,4]$

I thought of doing this $-5 \leq |\frac{df}{dx}(x)| \leq 5$ then $-5|dx| \leq |df| \leq 5|dx|$ , thus integrating all we get $-5x \leq f(x) \leq 5x$ so $f(1)$ is in $[-5,5]$ .

Also i fear that i never used the condition that $f(0)=0$ ?

Is this approach correct,any how can we play with these operators of $dx$ like this ,

thanks!

BAYMAX
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1 Answers1

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From $|f'(x)| \le 5$ it does not follow that $-5x \leq f(x) \leq 5x$ !

Example: $f(x)=5x+300123$.

By the mean value theorem there is $t \in (0,1)$ such that

$f(1)= \frac{f(1)-f(0)}{1-0}=f'(t)$, hence

$|f(1)| \le 5$, therefore $f(1) \in [-5,5]$

Fred
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  • Wow..that is nice,but i am curious that why we cannot do like that...,also $\frac{df}{dx}$ is a function of $x$ , any stress on that matter ! – BAYMAX Mar 02 '17 at 12:25