Continuous Functions are not Always Differentiable. But can we safely say that if a function f(x) is differentiable within range $(a,b)$ then it is continuous in the interval $[a,b]$ . If so , what is the logic behind it ?
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If the function is defined on $[a,b]$, then yes. It's a standard result, just open a calculus book to find a proof. – Git Gud Sep 14 '14 at 10:30
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Very good question. To start with, what do you know about the definition of differentiability and continuity? – Eric Auld Sep 14 '14 at 10:30
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A function is said to be differentiable if , it derivative is defined for every value in that range.And continuous definition , I am not sure about it. @EricAuld – Neer Sep 14 '14 at 10:37
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For continuity, it is (very informally): If $x$ is close to $y$, then $f(x)$ is close to $f(y)$. (Another informal way of putting it: If $h$ is very small, then $f(x+h)\approx f(x)$.) – Akiva Weinberger Sep 14 '14 at 10:55
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1For a more formal way of doing things: Let's say I want $f(x+h)$ and $f(x)$ to be "close": at most $\epsilon>0$ apart. Then, I can find a $\delta>0$ such that for all $|h|<\delta$, we have $|f(x+h)-f(x)|<\epsilon$. (This should be true for every $\epsilon$.) (Yes, this is a very confusing definition, sorry.) Another way of saying this is that $\lim_{a\to x}f(a)=f(x)$ or that $\lim_{h\to 0}f(x+h)-f(x)=0$. – Akiva Weinberger Sep 14 '14 at 11:03
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I understood ,thanks for sharing. @columbus8myhw – Neer Sep 14 '14 at 11:05
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@GitGud and others: Not so fast. Note open and closed intervals here! – Ted Shifrin Sep 14 '14 at 11:23
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1We cannot safely say that if $f$ is differentiable on $(a,b)$ then it is continuous on $[a,b]$! Just consider the function that's zero on $(0,1)$ and has value $1$ at $0$ and $1$. It is, however, necessarily continuous on $(a,b)$. – Najib Idrissi Sep 14 '14 at 11:23
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I will assume that $a<b$.
Consider the function $g: [a,b]\to {\mathbb R}$ which equals $0$ at $a$, and equals $1$ on the interval $(a,b]$. This function is differentiable on $(a,b)$ but is not continuous on $[a,b]$. Thus, "we can safely say..." is plain wrong.
However, one can define derivatives of an arbitrary function $f: [a,b]\to {\mathbb R}$ at the points $a$ and $b$ as $1$-sided limits: $$ f'(a):= \lim_{x\to a+} \frac{f(x)-f(a)}{x-a}, $$ $$ f'(b):= \lim_{x\to b-} \frac{f(x)-f(b)}{x-b}. $$ If these limits exist (as real numbers), then this function is called differentiable at the points $a, b$. For the points of $(a,b)$ the derivative is defined as usual, of course. The function $f$ is said to be differentiable on $[a,b]$ if its derivative exists at every point of $[a,b]$.
Now, the theorem is that a function differentiable on $[a,b]$ is also continuous on $[a,b]$. As for the proof, you can avoid $\epsilon$-$\delta$ definitions and just use limit theorems. For instance, to check continuity at $a$, use: $$ \lim_{x\to a+} (f(x)-f(a))= \lim_{x\to a+} (x-a) \lim_{x\to a+} \frac{f(x)-f(a)}{x-a} = 0\cdot f'(a)=0. $$ Hence, $$ \lim_{x\to a+} f(x)=f(a), $$ hence, $f$ is continuous at $a$. For other points the proof is the same.
Moishe Kohan
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Because a is arbitrary, there always exists b greater than a. Can't we say the fact that for all a f is continuous evreywhere?@MoisheKohan – Valentin May 07 '19 at 08:57
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@Val: I have no idea what you mean. I suggest you ask a new question, but make sure you include your thoughts about it. – Moishe Kohan May 07 '19 at 14:02
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The logic behind it lies in a little pool of definitions which, if you meditate on them, will start to make math come alive in a new way. Any good math textbook will take you there; my favorite is Spivak's Calculus.
Until then, intuitively, a function is continuous if its graph has no breaks, and differentiable if its graph has no corners and no breaks. So differentiability is stronger.
Eric Auld
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Let a point M(y,f(y)) that f is differentiable
For x $\ne$ y , f(x) - f(y) = $\frac{f(x) - f(y)}{x - y}$ * (x - y)
$\lim\limits_{x \to y}( {f(x) - f(y)} )$ = $\lim\limits_{x \to y} (\frac{f(x) - f(y)}{x - y} * (x - y))$ = f'(y) * 0 = 0
So $\lim\limits_{x \to y} {f(x) - f(y)}$ = 0. and $\lim\limits_{x \to y+} f(x)$ = f(y)
Hence f is continuous at y.
Now I will also prove that if a function is differentiable in [a,b] then f' is continuous on (a,b).
Let M(y,f(y)) be a point inside (a,b). In order for f to be differentiable at M,
$\lim\limits_{x \to y+} \frac{f(x) - f(y)}{x - y}$ =$\lim\limits_{x \to y-} \frac{f(x) - f(y)}{x - y}$ = $\lim\limits_{x \to y} \frac{f(x) - f(y)}{x - y}$ = f'(y).
Therefore, $\lim\limits_{x \to y+}{f'(x)}$ = $\lim\limits_{x \to y-}{f'(x)}$ = $\lim\limits_{x \to y}{f'(x)}$ = f'(y).
Hence, since M(y,f(y)) was chosen randomly f' is continuous in the (a,b) interval.
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A function is only differentiable on an open set, then it has no sense to say that your function is differentiable en $a$ or on $b$. But if $\lim_{x\to a^+}f'(x)$ and $\lim_{x\to b^-}f'(x)$ exists, then your function is $\mathcal C^1([a,b])$ and so yes your function is continuous on $[a,b]$. But this is stronger than just to check the continuity of $f$ on $a$ and on $b$.
idm
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Actually in the book it is written any arbitrary function f(x). No definite function is given. – Neer Sep 14 '14 at 11:10
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3It certainly makes sense for a function to be differentiable on, say, $[a,b]$. Just define $f'(a) = \lim_{x \to a^+} (f(x)-f(a))/(x-a)$ if the limit exists. It's in higher dimensions that things start to get complicated. – Najib Idrissi Sep 14 '14 at 11:22
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With your definition, it's just differentiable at the right of $a$ but not on $a$ ! – idm Sep 14 '14 at 11:34
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