   Chapter 2, Problem 30RE ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# Let g ( x ) = { 2 x − x 2     if   0 ≤ x ≤ 2 2 − x     if   2 < x ≤ 3 x − 4     if   3 < x < 4 π     if   x ≥ 4 a) For each of the numbers 2, 3, and 4, discover whether g is continuous from the left, continuous from the right, or continuous at the number.(b) Sketch the graph of g.

To determine

Whether g is continuous from the right, or from the left, or continuous at the numbers 2, 3 and 4.

Explanation

Given:

The function g(x)={2xx2if 0x22xif 2<x3x4if 3<x<4πif x4.

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:

Consider the piecewise function g(x)={2xx2if 0x22xif 2<x3x4if 3<x<4πif x4 .

Here, the function g(x)=2xx2 is a polynomial defined in the interval [0,2],.g(x)=2x is a polynomial defined in the interval (2,3], g(x)=x4 is a polynomial defined in the interval (3,4) and g(x)=π is a constant function defined in the interval [4,).

Since g(x)=2xx2, g(x)=2x and g(x)=x4 are polynomial functions and g(x)=π is a constant function and by theorem 1, the functions are continuous on its respective domains.

Therefore, g is continuous on the interval [0,2)(2,3)(3,4)(4,).

So that, g might be discontinuous at 2, 3 and 4.

At the number 2: Check the continuity of g.

At x=2, g(x)=2xx2.

This implies that, g(2)=0 is defined. (1)

The limit of the function g(x) as x approaches 2 is computed as follows.

Consider the left hand limit limx2g(x).

limx2g(x)=limx2(2xx2)=2(2)(2)2=44=0

Thus, the left-hand limit of the function as x approaches 2 is 0

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