Given $f:R\to R$ such that $f(0)=0$ and $|f'(x)|\le 5$ for all $x$. We can conclude that $f(1)$ is in
- $(5,6)$
- $[-5,5]$
- $(-\infty,-5)\cup (5,\infty)$
- $[-4,4]$
I could not solve it. Any help is appreciated thanks!
Asked
Active
Viewed 103 times
0
Departed
- 1,528
-
Wrong title, it seems. For the question in the body try $f=0$. – Dietrich Burde Dec 20 '16 at 20:26
-
yes @DietrichBurde, thanks. Yes f=0 rules out1 and 3, but not sure about 2 and 4? – Departed Dec 20 '16 at 20:29
-
finding answer by using different function is not what I am after? I want to learn the thing that should be used to find it? how should i do it subjectively? – Departed Dec 20 '16 at 20:32
3 Answers
10
It seems that $f$ is differentiable at $\Bbb R$, so we can use MVT in the interval $[0,1]$.
thus
$$\exists c\in(0,1) : f(1)-f(0)=(1-0)f'(c)$$
$$\implies f(1)=f'(c)$$
$$\implies |f(1)|=|f'(c)|\leq 5$$
$\implies f(1)\in[ -5, 5]$
hamam_Abdallah
- 62,951
3
Since $|f'(x)|\leq 5$ for all $x\in\Bbb R$, the most that $f$ can be increasing or decreasing by is $5$. That is, the largest rate of change of the function is when $f$ has a constant slope of $5$ (could be $-5$). So the largest range of values for $f(a)$, where $a\neq 0$, given $f(0)=0$ is $f(a)\in [-5a,5a]$, or $f(a)\in [5a,-5a]$ if $a<0$.
Dave
- 13,568
2
The other answers explain this very well, but I'd like to show an approach that works for this particular problem. Consider the function $f(x)=5x$. Then $f'(x)=5$ for all $x$, and $f(0)=0$, so it satisfies the given condition. Note, then, that $f(1)=5$. Then you can immediately deduce that (2) is the correct answer, as no other answer allows $f(1)=5$, even though $f(x)=5x$ is a valid function given these conditions.
I'll warn that this approach won't necessarily single out the answer for you. For example, if one of the answers was $[-6,6]$, you couldn't say that that one was wrong because it allows $f(1)=5$. In that case, you would have to use one of the other approaches mentioned to conclude that $[-5,5]$ is correct, and in general, you should use that kind of logic. More than anything, I think it's just useful to have a way to verify that your answer makes sense, if possible.
Kevin Long
- 5,159