   Chapter 2.5, Problem 41E

Chapter
Section
Textbook Problem

Find the numbers at which f is discontinuous. At which of these numbers is f continuous from the right, from the left, or neither? Sketch the graph of f. f ( x ) = { x 2   if   x < − 1 x       if   − 1 ≤ x < 1 1 / x   if   x ≥ 1

To determine

To find: The numbers at which the function f(x)={x2if x<1xif 1x<11xif x1 is discontinuous, continuous from the right, from the left, or neither. Sketch the graph of the function f(x)={x2if x<1xif 1x<11xif x1.

Explanation

Definition used: “A function f is continuous at a number a if limxaf(x)=f(a)”.

Note 1: “If f is defined near a, f is discontinuous at a whenever f is not continuous at a”.

Theorem used:

1. The functions such as “Polynomials, rational functions, root functions, trigonometric functions, inverse trigonometric functions, exponential functions and logarithmic functions” are continuous at every number in their domains.

2. A function f is continuous from the right at a number a if limxa+f(x)=f(a) and a function f is continuous from the left at a number a if limxaf(x)=f(a).

3. The limit limxaf(x)=L if and only if limxaf(x)=L=limxa+f(x).

Calculation:

By note 1, the function f is said to be discontinuous at x=a if anyone of the following conditions is not satisfied.

• f(a) is defined
• The limit of the function at the number a exists.
• limxaf(x)=f(a)

Consider the piecewise function f(x)={x2if x<1xif 1x<11xif x1 .

Here, the function f(x)=x2 is a polynomial function defined in the interval (,1), f(x)=x is a polynomial function defined in the interval (1,1) and f(x)=1x is a rational function defined in the interval (1,).

Since f(x)=x2, f(x)=x are polynomial functions and f(x)=1x is a rational function and by theorem 1, the functions are continuous on its respective domains.

Therefore, f is continuous on the interval (,1)(1,1)(1,).

So that, f might be discontinuous at −1 and 1.

Check the discontinuity of f at x=1.

At x=1, f(1)=1  is defined.

The limit of the function f(x) as x approaches a=1 is computed as follows

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