मराठी

Is Every Differentiable Function Continuous? - Mathematics

Advertisements
Advertisements

प्रश्न

Is every differentiable function continuous?

थोडक्यात उत्तर

उत्तर

Yes, if a function is differentiable at a point then it is necessary continuous at that point. 

\[\text{Proof : Let a function } f(x) \text { be differentiable at x = c . Then}, \]
\[ \lim_{x \to c} \frac{f(x) - f(c)}{x - c}\text { exists finitely } . \]
\[\text { Let } \lim_{x \to c} \frac{f(x) - f(c)}{x - c} = f'(c)\]
\[\text{In order to prove that f(x) is continous at x = c , it is sufficient to show that} \lim_{x \to c} f(x) = f(c)\]
\[ \lim_{x \to c} f(x) = \lim_{x \to c} \left\{ \left( \frac{f(x) - f(c)}{x - c} \right)\left( x - c \right) + f(c) \right\}\]
\[ \Rightarrow \lim_{x \to c} f(x) = \lim_{x \to c} \left[ \left\{ \frac{f(x) - f(c)}{x - c} \right\}\left( x - c \right) \right] + f(c)\]
\[ \Rightarrow \lim_{x \to c} f(x) = \lim_{x \to c} \left\{ \frac{f(x) - f(c)}{x - c} \right\} . \lim_{x \to c} \left( x - c \right) + f(c)\]
\[ \Rightarrow \lim_{x \to c} f(x) = f'(c) \times 0 + f(c)\]
\[ \Rightarrow \lim_{x \to c} f(x) = f(c)\]
\[\text{Hence, f(x) is continuous at x} = c .\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 10: Differentiability - Exercise 10.3 [पृष्ठ १७]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 10 Differentiability
Exercise 10.3 | Q 2 | पृष्ठ १७

व्हिडिओ ट्यूटोरियलVIEW ALL [4]

संबंधित प्रश्‍न

Examine the following function for continuity:

f(x) = | x – 5|


Discuss the continuity of the function f, where f is defined by `f(x) = {(-2,"," if x <= -1),(2x, "," if -1 < x <= 1),(2, "," if x > 1):}`


Discuss the continuity of the following functions at the indicated point(s): 

(i) \[f\left( x \right) = \begin{cases}\left| x \right| \cos\left( \frac{1}{x} \right), & x \neq 0 \\ 0 , & x = 0\end{cases}at x = 0\]

 


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

 


Discuss the continuity of \[f\left( x \right) = \begin{cases}2x - 1 & , x < 0 \\ 2x + 1 & , x \geq 0\end{cases} at x = 0\]


Determine the value of the constant k so that the function

\[f\left( x \right) = \begin{cases}k x^2 , if & x \leq 2 \\ 3 , if & x > 2\end{cases}\text{is continuous at x} = 2 .\]


If the functions f(x), defined below is continuous at x = 0, find the value of k. \[f\left( x \right) = \begin{cases}\frac{1 - \cos 2x}{2 x^2}, & x < 0 \\ k , & x = 0 \\ \frac{x}{\left| x \right|} , & x > 0\end{cases}\] 

 


Find the points of discontinuity, if any, of the following functions: 

\[f\left( x \right) = \begin{cases}\frac{e^x - 1}{\log_e (1 + 2x)}, & \text{ if }x \neq 0 \\ 7 , & \text{ if } x = 0\end{cases}\]

Determine if \[f\left( x \right) = \begin{cases}x^2 \sin\frac{1}{x} , & x \neq 0 \\ 0 , & x = 0\end{cases}\] is a continuous function?

 


Define continuity of a function at a point.

 

The function 

\[f\left( x \right) = \frac{4 - x^2}{4x - x^3}\]

 


Let f (x) = | x | + | x − 1|, then


If \[f\left( x \right) = \begin{cases}mx + 1 , & x \leq \frac{\pi}{2} \\ \sin x + n, & x > \frac{\pi}{2}\end{cases}\] is continuous at \[x = \frac{\pi}{2}\]  , then

 


The value of f (0) so that the function 

\[f\left( x \right) = \frac{2 - \left( 256 - 7x \right)^{1/8}}{\left( 5x + 32 \right)^{1/5} - 2},\]  0 is continuous everywhere, is given by


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

 


If  \[f\left( x \right) = \begin{cases}\frac{1 - \sin^2 x}{3 \cos^2 x} , & x < \frac{\pi}{2} \\ a , & x = \frac{\pi}{2} \\ \frac{b\left( 1 - \sin x \right)}{\left( \pi - 2x \right)^2}, & x > \frac{\pi}{2}\end{cases}\]. Then, f (x) is continuous at  \[x = \frac{\pi}{2}\], if

 


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


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


Discuss the continuity and differentiability of 

\[f\left( x \right) = \begin{cases}\left( x - c \right) \cos \left( \frac{1}{x - c} \right), & x \neq c \\ 0 , & x = c\end{cases}\]

The function f (x) = e|x| is


Let f (x) = |sin x|. Then,


Let \[f\left( x \right) = \begin{cases}1 , & x \leq - 1 \\ \left| x \right|, & - 1 < x < 1 \\ 0 , & x \geq 1\end{cases}\] Then, f is 


Examine the continuity of f(x)=`x^2-x+9  "for"  x<=3`

=`4x+3  "for"  x>3,  "at"  x=3` 


Discuss continuity of f(x) =`(x^3-64)/(sqrt(x^2+9)-5)` For x ≠ 4 

= 10 for x = 4  at x = 4


Find the value of k for which the function f (x ) =  \[\binom{\frac{x^2 + 3x - 10}{x - 2}, x \neq 2}{ k , x^2 }\] is continuous at x = 2 .

 
 

Find k, if the function f is continuous at x = 0, where

`f(x)=[(e^x - 1)(sinx)]/x^2`,      for x ≠ 0

     = k                             ,        for x = 0


Find `dy/dx if y = tan^-1 ((6x)/[ 1 - 5x^2])`


 If the function f is continuous at x = I, then find f(1), where f(x) = `(x^2 - 3x + 2)/(x - 1),` for x ≠ 1


If the function f is continuous at x = 2, then find 'k' where

f(x) = `(x^2 + 5)/(x - 1),` for  1< x ≤ 2 
      = kx + 1 , for x > 2


Find the value of the constant k so that the function f defined below is continuous at x = 0, where f(x) = `{{:((1 - cos4x)/(8x^2)",", x ≠ 0),("k"",", x = 0):}`


If f(x) = `{{:((x^3 + x^2 - 16x + 20)/(x - 2)^2",", x ≠ 2),("k"",", x = 2):}` is continuous at x = 2, find the value of k.


The number of points at which the function f(x) = `1/(log|x|)` is discontinuous is ______.


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


Given the function f(x) = `1/(x + 2)`. Find the points of discontinuity of the composite function y = f(f(x))


Examine the differentiability of f, where f is defined by
f(x) = `{{:(x^2 sin  1/x",",  "if"  x ≠ 0),(0",", "if"  x = 0):}` at x = 0


Find the values of p and q so that f(x) = `{{:(x^2 + 3x + "p"",",  "if"  x ≤ 1),("q"x + 2",",  "if"  x > 1):}` is differentiable at x = 1


If the following function is continuous at x = 2 then the value of k will be ______.

f(x) = `{{:(2x + 1",", if x < 2),(                 k",", if x = 2),(3x - 1",", if x > 2):}`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×