# First derivative test in Spivak Calculus used without proof?

So I'm reading Spivak calculus and I noticed two things:

1. The first derivative test is stated and used without proof.

2. When discussing increasing functions, he presented the following:

So by using the mean value theorem, this says if $$f$$ is continuous on $$[a,b]$$ and $$f'>0$$ for all $$x$$ in $$[a,b]$$ then $$f$$ is increasing on $$[a,b]$$

However, according to my understanding the mean value theorem asserts a slightly stronger statement, that is $$f'$$ needs only to be $$>0$$ in $$(a,b)$$ that is, excluding the end points, and even in this case $$f$$ will be increasing on $$[a,b]$$

And this detail is needed when proving the first derivative test, for example as in here. Am I correct?

Edit

So I'm basically asking what is the proof of the first derivative test, and the part that confuses me is proving the following:

If $$f$$ is (for instance) increasing on $$(a,b)$$ and continuous at $$b$$ then it is increasing on $$(a,b]$$

• His proof is fine, and your slight generalization (with the necessary continuity hypothesis of $f$ at $a$ and $b$ like in your link) can be easily deduced from his statement. Commented Nov 16, 2023 at 13:49
• @AnneBauval indeed I forget to mention the continuity hypothesis of $f$ at $a$ and $b$, thanks for noting that. However how does this generalization follow from his statement? Also this generalization is essential to prove the first derivative test, right? Commented Nov 16, 2023 at 13:57
• Either directly in the proof of the first derivative test, or in the proof of your corollary, you can just say: if $f$ is (for instance) increasing on $(a,b)$ and continuous at $b$ then it is increasing on $(a,b].$ Commented Nov 16, 2023 at 14:00
• yes, how can we prove this intuitive statement? Commented Nov 16, 2023 at 14:09
• Not "intuitive": For every $x\in(a,b),$ let $y\in(x,b).$ Then, $f(b)=\lim_{t\to b\atop y<t<b}f(t)\ge f(y)>f(x).$ Commented Nov 16, 2023 at 14:23

If $$f$$ is increasing on $$(a,b)$$ and continuous at $$b$$ then it is increasing on $$(a,b].$$
Let $$f$$ be increasing on $$(a,b)$$ and continuous at $$b$$, and let $$a We shall prove that $$f(x) If $$y this holds by hypothesis. So, we may assume $$y=b,$$ and choose some $$z\in(x,b).$$ Then, since $$\forall t\in(z,b),\;f(t)>f(z):$$ $$f(b)=\lim_{t\to b\atop z hence $$f(b)>f(x),$$ q.e.d.
• why $f(t) \geq f(z)$, not $f(t) > f(z)$? Also is this a one sided limit version of half-the squeeze theorem? if so what is its proof? Commented Nov 19, 2023 at 16:13
• 1) Although $f(t)\ge f(z)$ would be sufficient for the consequence about the limit, what I wrote is $f(t)>f(z)$ ($\forall t\in(z,b)$). 2) The consequence used here is simpler than "a one sided limit version of half-the squeeze theorem", whatever that means: it is the fact that whenever a quantity stays $\ge$ some constant $C,$ its limit (if it exists) is also $\ge C.$ It is a good exercise to prove it (directly from the epsilon-delta definition of the limit). Note that even wen the "quantity" stays $>C,$ the limit may be $=C.$ Commented Nov 19, 2023 at 18:35
• I meant $f(t) \geq f(z)$ after the limit, your last sentence in your comment clarify why, thanks a lot Commented Nov 19, 2023 at 18:44