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Let $f:[a, b] \rightarrow \mathbb{R}$ where $a, b \in \mathbb{R}$ with $a <b$. Suppose $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$. Suppose $f$ is monotonically increasing on $[a,b]$. Show that $f'(x) \ge 0$ for all $x \in (a,b)$.

My attempt: I tried using the Mean Value Theorem, but it doesn't quite seem to work. For example, by the MVT we can conclude that there exists a $c \in (a,b)$ such that $f(b) - f(a) = f'(c) (b-a)$. Which implies that $f'(c) = \frac{f(b) - f(a)}{b-a}$. Now since $f$ is monotonically increasing, $f(b) - f(a) \ge 0$ whenever $b>a$, so $f'(c) \ge 0$. But this only shows for one particular $c \in (a,b)$, and the question asks to show this is true for ALL $x \in (a,b)$. What can I do to complete the proof?

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    $\begingroup$ My attempt: I tried using the Mean Value Theorem, but it doesn't quite seem to work. May I ask, then, why you chose to put "using mean value theorem" in the title? Maybe MVT is indeed not the right way to solve this, and such a title can only discourage better alternative answers. $\endgroup$
    – dxiv
    Aug 8, 2017 at 3:52

4 Answers 4

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For $x \in (a,b)$ $$f'(x)=\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}\geq0$$

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  • $\begingroup$ While this is true, and is indeed the standard way to prove the stated result, it doesn't seem to use the Mean Value Theorem, which is a stated desideratum of the proof. $\endgroup$ Aug 8, 2017 at 2:12
  • $\begingroup$ Thanks @Shahiba Arora, but is there a proof using MVT? $\endgroup$
    – user40333
    Aug 8, 2017 at 2:12
  • $\begingroup$ No use of MVT at all!! $\endgroup$
    – Math Lover
    Aug 8, 2017 at 2:15
  • $\begingroup$ @JohnColeman$:$ Mean value theorem is used to prove the converse. $\endgroup$
    – Anacardium
    Mar 9 at 5:29
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Your approach using MVT will never guarantee the location of $c$ or may not cover all $c\in(a, b)$. Consider for example $f(x) = x^{3}$ in $[-1,1]$. Since $f$ is strictly increasing it follows that the ratio $(f(b) - f(a)) /(b-a) >0$ for any two distinct points $a, b\in[-1,1]$ and thus the corresponding $c$ guaranteed by MVT will always have $f'(c) >0$. This misses the point $c=0$ where derivative vanishes. Thus your proof can not be salvaged. Even isolating such troublesome points like the one given in above example and then showing that derivative vanishes there is a bit difficult.

The proper and much simpler approach is to use the definition of derivative as given in Sahiba Arora's answer.

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Hint: For a given $x$, there exists $a \le c \le x \le d \le b$ such that $$\frac{f(d)-f(c)}{d-c}=f'(x).$$

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  • $\begingroup$ But that isn't what the Mean Value Theorem itself says. $\endgroup$ Aug 8, 2017 at 1:59
  • $\begingroup$ @JohnColeman The idea is to consider a 'new' function defined on $(c,d)=(x-\delta,x+\eta)$ such that the mean-value theorem gives the desired result. $\endgroup$
    – Math Lover
    Aug 8, 2017 at 2:03
  • $\begingroup$ How would you finish your proof @Math Lover? $\endgroup$
    – user40333
    Aug 8, 2017 at 2:09
  • $\begingroup$ @user40333 For any given $x \in (a,b)$, we consider the function defined on $[y,z]$, where $x \in (y,z)$, and $a \le y < x < z \le b$ such that $\frac{f(z)-f(y)}{z-y}=f'(x)$. Since $f(z) \ge f(y)$, $f'(x) \ge 0$. $\endgroup$
    – Math Lover
    Aug 8, 2017 at 2:13
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    $\begingroup$ The problem is that unless I am missing something your statement is (in general) false. The converse of the Mean Value Theorem doesn't hold without some additional assumptions. For example $f(x) = x^3$ satisfies $f'(0) = 0$ but $\frac{f(d)-f(c)}{d-c}$ never equals $0$ for any $c,d$ which straddle $0$. Now it might be the case that $f$ increasing is enough to guarantee the result, but if so, that would be a result which is harder to show than the result being proved. $\endgroup$ Aug 8, 2017 at 2:22
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Your proof can be made to work if $f'$ is continuous. Let $x∈ (a,b)$. Then for each $n\gg 1$, apply MVT as you have on the interval $(x-\frac1n,x+\frac1n)$ to discover an $x_n$ with $|x_n - x|<1/n$ and $f'(x_n)\geq 0$. Just take $n→∞$ to conclude, using continuity of $f'$.

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