Question

$\Large {\sf{\pmb{\underline{\underline{Challenge...}}}}}$

$\sf \circ\ {Find\ all\ points\ of\ discontinuous\ of\ f,\ where\ f\ is\ defined\ by,}$

$\sf f(x)\ = \begin{cases} &\sf{2x\ +\ 3,\: \: \: \: \: \: \: \: if\ x \leq 2} \\ \\ &\sf{2x\ -\ 3,\: \: \: \: \: \: \: \: if\ x > 2} \end{cases}$

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Answer 1

The Given Function is:-

$\sf f(x)\ = \begin{cases} &\sf{2x\ +\ 3,\: \: \: \: \: \: \: \: if\ x \leq 2} \\ \\ &\sf{2x\ -\ 3,\: \: \: \: \: \: \: \: if\ x > 2} \end{cases}$

Explaination:-

• When x < 1 , f(x) = x+2 which being polynomial is continuous for all x < 1 

• When x > 1 f(x) = x - 2 ,which being polynomial is continuous for all x > 1 

Now we consider point x= 2

At x = 2 

L.H.L = lim x→2- f(x) = lim x→2- f(2x+3) 

= lim h→0 { 2(2-h) +3 }

= 7

Now R.H.L 

= lim x→2+ f(x) = lim x→2+ f(2x-3) 

= lim h→ 0 2(2-h) -3 

= 1 

Thus L.H.L ≠ R.H.L 

hence , Function is discontinuous at x = 2

Answer 2

Basic Definition of Continuity :-

A function f(x) is said to be continuous at x = a, iff

$\underline{\boxed{\displaystyle\tt \:\lim_{x\to \: a^-} \: = \: \tt \:\lim_{x\to \: a^+} = f(a)}}$

$\large\underline{\sf{Solution-}}$

Given that,

$\rm \: f(x)\ = \begin{cases} &\bf{2x\ +\ 3,\: \: \: \: \: \: \: \: if\ x \leq 2} \\ \\ &\bf{2x\ -\ 3,\: \: \: \: \: \: \: \: if\ x > 2} \end{cases}$

Here,

• Breaking point is x = 2.

So,

• We have to check the continuity of the function at x = 2.

Step :- 1

Value of f(x) at x = 2

$\rm :\longmapsto\:f(2) = 2 \times 2 + 3 = 4 + 3 = 7 - - (1)$

Step :- 2

• Right Hand Limit at x = 2

$\rm :\longmapsto\:\displaystyle\tt \:\lim_{x\to2^ + } \:f(x)$

$\rm \: \: = \: \: \: \displaystyle\tt \:\lim_{x\to2^ + } \:(2x - 3)$

$\bigg\{ \red{ \bf \: Put \: x \: = \: 2 + h} \\ \red{\rm \: As \: x \to \: 2 \implies \: h \to \: 0} \bigg\}$

$\rm \: \: = \: \: \: \displaystyle\tt \:\lim_{h\to0} \:(2(2 + h) - 3)$

$\rm \: \: = \: \: \: \displaystyle\tt \:\lim_{h\to0} \:(4 + 2h - 3)$

$\rm \: \: = \: \: \: \displaystyle\tt \:\lim_{h\to0} \:(1 + 2h)$

$\rm \: \: = \: \: \: 1 - - - (2)$

From equation (1) and equation (2), we concluded that

$\rm :\longmapsto\:f(2) \: \ne \: \: \tt \:\lim_{x \: \to \: 2^ + }$

$\bf\implies \:f(x) \: is \: not \: continuos \: at \: x = 2$

$\bf\implies \:only \: one \: point \: of \: discontinuity.$

Additional Information :-

1. If f and g are two continuous function at x = a, then

• f + g is continuous at x = a.

• f - g is continuous at x = a.

• f×g is continuous at x = a

• fog is continuous at x = a.

2. Every differentiable function is always continuous but continuous function may or may not be differentiable.

3 Every sine, cosine function is always continuous on R.