मराठी

Examine the Continuity of the Function F ( X ) = { 3 X − 2 , X ≤ 0 X + 1 , X > 0 a T X = 0 Also Sketch the Graph of this Function.

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प्रश्न

Examine the continuity of the function  

\[f\left( x \right) = \left\{ \begin{array}{l}3x - 2, & x \leq 0 \\ x + 1 , & x > 0\end{array}at x = 0 \right.\]

Also sketch the graph of this function.

बेरीज
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उत्तर

Meaning of continuity of function - If we talk about a general meaning of continuity of a function f(x) , we can say that if we plot the coordinates (x, f(x)) and try to join all those points in the specified region, we can do so without picking our pen i.e you will put your pen/pencil on graph paper and you can draw the curve without any breakage.

Mathematically we define the same thing as given below:

A function f(x) is said to be continuous at x = c where c is x-coordinate of the point at which continuity is to be checked

If: 

`lim_(h->0)f(c - h) = lim_(h->0)f(c + h) = f(c)`

where h is a very small positive no (can assume h = 0.00000000001 like this)

It means :- 

Limiting value of the left neighbourhood of x = c also called left hand limit LHL `{i.e lim_(h->0)f(c - h)}` must be equal to limiting value of right neighbourhood of x= c called right hand limit RHL `{i.e lim_(h->0)f(c + h)}` and both must be equal to the value of f(x) at x = c i.e. f(c). 

Thus, it is the necessary condition for a function to be continuous.

So, whenever we check continuity we try to check above equality if it holds true, function is continuous else it is discontinuous.

Lets solve now:

Given function is

f(x) = `{(3x - 2, if x ≤ 0),(x + 1, if x > 0):}`         ...(2)

We need to check whether f(x) is continuous at x=0 or not

For this we need to check LHL, RHL and value of function at x = 0

Clearly,

f(0) = 3*0 - 2 = -2 [from equation 2] 

LHL = `lim_(h->0)f(0 - h) = lim_(h->0)f(-h) = lim_(h->0){3(-h)-2} = -2`

RHL = `lim_(h->0)f(0 + h) = lim_(h->0)f(h) = lim_(h->0){h + 1} = 0 + 1 = 1`

As, LHL ≠ RHL

f(x) is discontinuous at x = 0

This can also be proved by plotting f(x) on cartesian plane.

For x >0, we need to plot

y = x + 1

put y = 0, we get x = -1 and for second point we put x = 0 and thus get y=1

two points are enough to plot the straight line.

Two coordinates are (-1,0) and (0,1)

For x ≤ 0, we need to plot 

y = 3x - 2

put x = 0 then y = -2

on putting y = 0 we get x 2/3

two coordinates are (0, -2) and `(2/3, 0)`

Graph:

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पाठ 8: Continuity - Exercise 9.1 [पृष्ठ १८]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 8 Continuity
Exercise 9.1 | Q 14 | पृष्ठ १८

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