English

Is Every Continuous Function Differentiable? - Mathematics

Advertisements
Advertisements

Question

Is every continuous function differentiable?

Advertisements

Solution

No, function may be continuous at a point but may not be differentiable at that point .
For example: function  

\[f(x) = |x|\]  is continuous at 
 
\[x = 0\]  but it is not differentiable at 
\[x = 0\]
shaalaa.com
  Is there an error in this question or solution?
Chapter 10: Differentiability - Exercise 10.3 [Page 17]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 10 Differentiability
Exercise 10.3 | Q 3 | Page 17

RELATED QUESTIONS

Find the value of 'k' if the function

`f(X)=(tan7x)/(2x) ,  "for " x != 0 `

`=k`,            for x=0

is continuos at x=0


Examine the following function for continuity:

f(x) = `1/(x - 5)`, x ≠ 5


Show that 

\[f\left( x \right) = \begin{cases}\frac{\left| x - a \right|}{x - a}, when & x \neq a \\ 1 , when & x = a\end{cases}\] is discontinuous at x = a.

Show that 

\[f\left( x \right) = \begin{cases}\frac{\sin 3x}{\tan 2x} , if x < 0 \\ \frac{3}{2} , if x = 0 \\ \frac{\log(1 + 3x)}{e^{2x} - 1} , if x > 0\end{cases}\text{is continuous at} x = 0\]


Discuss the continuity of the function f(x) at the point x = 0, where  \[f\left( x \right) = \begin{cases}x, x > 0 \\ 1, x = 0 \\ - x, x < 0\end{cases}\]

 


If \[f\left( x \right) = \begin{cases}\frac{x - 4}{\left| x - 4 \right|} + a, \text{ if }  & x < 4 \\ a + b , \text{ if } & x = 4 \\ \frac{x - 4}{\left| x - 4 \right|} + b, \text{ if } & x > 4\end{cases}\]  is continuous at x = 4, find ab.

 


Find the value of k for which \[f\left( x \right) = \begin{cases}\frac{1 - \cos 4x}{8 x^2}, \text{ when}  & x \neq 0 \\ k ,\text{ when }  & x = 0\end{cases}\] is continuous at x = 0;

 


In each of the following, find the value of the constant k so that the given function is continuous at the indicated point;  

\[f\left( x \right) = \begin{cases}k( x^2 - 2x), \text{ if }  & x < 0 \\ \cos x, \text{ if }  & x \geq 0\end{cases}\] at x = 0

In each of the following, find the value of the constant k so that the given function is continuous at the indicated point; 

\[f\left( x \right) = \begin{cases}kx + 1, \text{ if }  & x \leq \pi \\ \cos x, \text{ if }  & x > \pi\end{cases}\] at x = π

Find the points of discontinuity, if any, of the following functions:  \[f\left( x \right) = \begin{cases}\frac{\sin 3x}{x}, & \text{ if }   x \neq 0 \\ 4 , & \text{ if }  x = 0\end{cases}\]

 


If the function \[f\left( x \right) = \begin{cases}\left( \cos x \right)^{1/x} , & x \neq 0 \\ k , & x = 0\end{cases}\] is continuous at x = 0, then the value of k is


Let f (x) = | x | + | x − 1|, then


\[f\left( x \right) = \begin{cases}\frac{\sqrt{1 + px} - \sqrt{1 - px}}{x}, & - 1 \leq x < 0 \\ \frac{2x + 1}{x - 2} , & 0 \leq x \leq 1\end{cases}\]is continuous in the interval [−1, 1], then p is equal to

 


If  \[f\left( x \right) = \begin{cases}a \sin\frac{\pi}{2}\left( x + 1 \right), & x \leq 0 \\ \frac{\tan x - \sin x}{x^3}, & x > 0\end{cases}\] is continuous at x = 0, then a equals


If  \[f\left( x \right) = \left\{ \begin{array}a x^2 + b , & 0 \leq x < 1 \\ 4 , & x = 1 \\ x + 3 , & 1 < x \leq 2\end{array}, \right.\] then the value of (ab) for which f (x) cannot be continuous at x = 1, is

 


If  \[f\left( x \right) = \begin{cases}\frac{\sin \left( \cos x \right) - \cos x}{\left( \pi - 2x \right)^2}, & x \neq \frac{\pi}{2} \\ k , & x = \frac{\pi}{2}\end{cases}\]is continuous at x = π/2, then k is equal to


Show that f(x) = x1/3 is not differentiable at x = 0.


Show that \[f\left( x \right) =\]`{(12x, -,13, if , x≤3),(2x^2, +,5, if x,>3):}` is differentiable at x = 3. Also, find f'(3).


Show that the function 

\[f\left( x \right) = \begin{cases}x^m \sin\left( \frac{1}{x} \right) &, x \neq 0 \\ 0 &, x = 0\end{cases}\]

(i) differentiable at x = 0, if m > 1
(ii) continuous but not differentiable at x = 0, if 0 < m < 1
(iii) neither continuous nor differentiable, if m ≤ 0


Discuss the continuity and differentiability of f (x) = e|x| .


Write the points where f (x) = |loge x| is not differentiable.


Write the points of non-differentiability of 

\[f \left( x \right) = \left| \log \left| x \right| \right| .\]

If \[f\left( x \right) = \sqrt{1 - \sqrt{1 - x^2}},\text{ then } f \left( x \right)\text {  is }\] 


If \[f\left( x \right) = x^2 + \frac{x^2}{1 + x^2} + \frac{x^2}{\left( 1 + x^2 \right)} + . . . + \frac{x^2}{\left( 1 + x^2 \right)} + . . . . ,\] 

then at x = 0, f (x)


Let f (x) = |sin x|. Then,


The total cost C for producing x units is Rs (x2 + 60x + 50) and the price is Rs (180 - x) per unit. For how many units the profit is maximum.


Find the value of 'k' if the function 
f(x) = `(tan 7x)/(2x)`,                   for x ≠ 0.
      = k                                        for x = 0.
is continuous at x = 0.


 If the function f (x) = `(15^x - 3^x - 5^x + 1)/(x tanx)`,  x ≠ 0 is continuous at x = 0 , then find f(0).


Examine the continuity of the followin function : 

  `{:(,f(x),=x^2cos(1/x),",","for "x!=0),(,,=0,",","for "x=0):}}" at "x=0`   


Show that the function f given by f(x) = `{{:(("e"^(1/x) - 1)/("e"^(1/x) + 1)",", "if"  x ≠ 0),(0",",  "if"  x = 0):}` is discontinuous at x = 0.


The number of points at which the function f(x) = `1/(x - [x])` is not continuous is ______.


Examine the continuity of the function f(x) = x3 + 2x2 – 1 at x = 1


f(x) = `{{:((2x^2 - 3x - 2)/(x - 2)",", "if"  x ≠ 2),(5",", "if"  x = 2):}` at x = 2


f(x) = `{{:(("e"^(1/x))/(1 + "e"^(1/x))",", "if"  x ≠ 0),(0",", "if"  x = 0):}` at x = 0 


Prove that the function f defined by 
f(x) = `{{:(x/(|x| + 2x^2)",",  x ≠ 0),("k",  x = 0):}`
remains discontinuous at x = 0, regardless the choice of k.


A function f: R → R satisfies the equation f( x + y) = f(x) f(y) for all x, y ∈ R, f(x) ≠ 0. Suppose that the function is differentiable at x = 0 and f′(0) = 2. Prove that f′(x) = 2f(x).


If f is continuous on its domain D, then |f| is also continuous on D.


The composition of two continuous function is a continuous function.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×