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Discuss the continuity of the function f, where f is defined by: f(x) = {โˆ’2, if ๐‘ฅ โ‰ค โˆ’1, 2โข๐‘ฅ, if โˆ’1 < ๐‘ฅ โ‰ค 1, 2, if ๐‘ฅ > 1 - Mathematics

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Question

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):}`

Sum
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Solution

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

f(x) = −2 for x < 1;

−1 < x < 1, f(x) = 2x and

x > 1, f(x) = 2 is a polynomial function.

So this is a function.

At x = −1,

`lim_(x -> 1^-)` f(x) = `lim_(x -> 1^-)` (−2) = −2

`lim_(x -> 1^+)` f(x) = `lim_(x -> 1^+)` (2x)

= `lim_(h -> 0)` [2 (−1 + h)]

= `lim_(h -> 0)` (−2 + 2h)

= −2 + 0

= −2

f(−1) = −2

Hence, f is continuous at x = −1.

At x = 1,

`lim_(x -> 1^-)` f(x) = `lim_(x -> 1^-)` (2x)

= `lim_(h -> 0)` [2(1 − h)]

= `lim_(h -> 0)` (2 − 2h)

= 2 − 2 × 0

= 2

`lim_(x -> 1^+)` f(x) = `lim_(x -> 1^+)` (2) = 2

f(1) = 2 × 1 = 2

Hence, f is continuous at x = 1.

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Chapter 5: Continuity and Differentiability - Exercise 5.1 [Page 160]

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NCERT Mathematics Part 1 and 2 [English] Class 12
Chapter 5 Continuity and Differentiability
Exercise 5.1 | Q 16 | Page 160

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