
Say true or false:
Every continuous function is differentiable.
Answer
565.5k+ views
Hint: We will first write the fact that every differentiable function is continuous and then see if the converse is true or not. We will then just try to disprove the statement using any example which doesn’t follow the given rule.
Complete step by step answer:
We have the statement which is given to us in the question that: Every continuous function is differentiable.
Since, we know that “every differentiable function is always continuous”.
We just now need to check if the converse is also true or not.
Let us take the example of $f(x) = |x|$.
If forms a pointed edge at x = 0.
We will check its differentiability at 0 only.
We know that a function is differentiable at c if $f'(c) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(c + h) - f(c)}}{h}$ exists.
Let us check about $f(x) = |x|$ at x = 0.
$ \Rightarrow f'(0) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(0 + h) - f(0)}}{h}$
Now, putting in the function, we will get:-
$ \Rightarrow f'(0) = \mathop {\lim }\limits_{h \to 0} \dfrac{{|0 + h| - |0|}}{h}$
We can rewrite it as:
$ \Rightarrow f'(0) = \mathop {\lim }\limits_{h \to 0} \dfrac{{|h|}}{h}$
Now, it can be either equal to 1 or -1 depending on if h is approaching 0 from the right of the number line or left respectively. Therefore, the limits do not exist and thus the function is not differentiable.
But we see that $f(x) = |x|$ is continuous because $\mathop {\lim }\limits_{x \to c} f(x) = \mathop {\lim }\limits_{x \to c} |x| = f(c)$ exists for all the possible values of c.
Therefore, the given statement is false.
Note:
The students must note that “Every differentiable function is continuous”. We use this fact in a lot of questions. So, let us prove this to know the reason behind it.
Let us say we have a function f(x) which is differentiable at x = c.
So, by using the definition of differentiability, we will get:-
$f'(c) = \mathop {\lim }\limits_{x \to c} \dfrac{{f(x) - f(c)}}{{x - c}}$
We can rewrite it as:
$ \Rightarrow \mathop {\lim }\limits_{x \to c} f(x) - f(c) = \mathop {\lim }\limits_{x \to c} f'(c) \times (x - c)$
Rewriting it again as follows:
$ \Rightarrow \mathop {\lim }\limits_{x \to c} f(x) - f(c) = f'(c)\mathop {\lim }\limits_{x \to c} (x - c)$
Simplifying the RHS, we will then obtain:-
$ \Rightarrow \mathop {\lim }\limits_{x \to c} f(x) - f(c) = f'(c) \times 0$
Simplifying the RHS further, we will then obtain:-
$ \Rightarrow \mathop {\lim }\limits_{x \to c} f(x) - f(c) = 0$
Therefore, we have:-
$ \Rightarrow \mathop {\lim }\limits_{x \to c} f(x) = f(c)$
Hence, the function is continuous at x = c.
Therefore, it is proved that “Every differentiable function is continuous”.
Complete step by step answer:
We have the statement which is given to us in the question that: Every continuous function is differentiable.
Since, we know that “every differentiable function is always continuous”.
We just now need to check if the converse is also true or not.
Let us take the example of $f(x) = |x|$.
If forms a pointed edge at x = 0.
We will check its differentiability at 0 only.
We know that a function is differentiable at c if $f'(c) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(c + h) - f(c)}}{h}$ exists.
Let us check about $f(x) = |x|$ at x = 0.
$ \Rightarrow f'(0) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(0 + h) - f(0)}}{h}$
Now, putting in the function, we will get:-
$ \Rightarrow f'(0) = \mathop {\lim }\limits_{h \to 0} \dfrac{{|0 + h| - |0|}}{h}$
We can rewrite it as:
$ \Rightarrow f'(0) = \mathop {\lim }\limits_{h \to 0} \dfrac{{|h|}}{h}$
Now, it can be either equal to 1 or -1 depending on if h is approaching 0 from the right of the number line or left respectively. Therefore, the limits do not exist and thus the function is not differentiable.
But we see that $f(x) = |x|$ is continuous because $\mathop {\lim }\limits_{x \to c} f(x) = \mathop {\lim }\limits_{x \to c} |x| = f(c)$ exists for all the possible values of c.
Therefore, the given statement is false.
Note:
The students must note that “Every differentiable function is continuous”. We use this fact in a lot of questions. So, let us prove this to know the reason behind it.
Let us say we have a function f(x) which is differentiable at x = c.
So, by using the definition of differentiability, we will get:-
$f'(c) = \mathop {\lim }\limits_{x \to c} \dfrac{{f(x) - f(c)}}{{x - c}}$
We can rewrite it as:
$ \Rightarrow \mathop {\lim }\limits_{x \to c} f(x) - f(c) = \mathop {\lim }\limits_{x \to c} f'(c) \times (x - c)$
Rewriting it again as follows:
$ \Rightarrow \mathop {\lim }\limits_{x \to c} f(x) - f(c) = f'(c)\mathop {\lim }\limits_{x \to c} (x - c)$
Simplifying the RHS, we will then obtain:-
$ \Rightarrow \mathop {\lim }\limits_{x \to c} f(x) - f(c) = f'(c) \times 0$
Simplifying the RHS further, we will then obtain:-
$ \Rightarrow \mathop {\lim }\limits_{x \to c} f(x) - f(c) = 0$
Therefore, we have:-
$ \Rightarrow \mathop {\lim }\limits_{x \to c} f(x) = f(c)$
Hence, the function is continuous at x = c.
Therefore, it is proved that “Every differentiable function is continuous”.
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