Question 30 of 33intermediate🔧 ApplyShort Answer3 marks

Let f,g:R→Rf, g : R \rightarrow R be defined, respectively by f(x)=x+1f(x) = x + 1, g(x)=2x−3g(x) = 2x - 3. Find f+g,f−gf + g, f - g and fg\frac{f}{g}.

Correct Answer

Given f(x)=x+1f(x) = x + 1 and g(x)=2x−3g(x) = 2x - 3, we find the required functions using the definitions of algebra of real functions.

(f+g)(x)(f + g)(x): By definition, (f+g)(x)=f(x)+g(x)=(x+1)+(2x−3)=3x−2(f + g)(x) = f(x) + g(x) = (x + 1) + (2x - 3) = 3x - 2.

(f−g)(x)(f - g)(x): By definition, (f−g)(x)=f(x)−g(x)=(x+1)−(2x−3)=−x+4(f - g)(x) = f(x) - g(x) = (x + 1) - (2x - 3) = -x + 4.

(fg)(x)\left( \frac{f}{g} \right)(x): By definition, (fg)(x)=f(x)g(x)=x+12x−3\left( \frac{f}{g} \right)(x) = \frac{f(x)}{g(x)} = \frac{x + 1}{2x - 3}, provided g(x)≠0g(x) \neq 0. Since g(x)=2x−3=0g(x) = 2x - 3 = 0 when x=32x = \frac{3}{2}, we have (fg)(x)=x+12x−3,x≠32\left( \frac{f}{g} \right)(x) = \frac{x + 1}{2x - 3}, x \neq \frac{3}{2}.

Exercise: Miscellaneous Exercise on Chapter 2 | Q: 7 | (Page 17)
For More Understanding

Explanation

This question tests the algebra of real functions as defined in Section 2.4.2 of the textbook. Students must apply the definitions: (f + g)(x) = f(x) + g(x), (f - g)(x) = f(x) - g(x), and (f/g)(x) = f(x)/g(x) with the condition g(x) ≠\neq 0. The solution follows the pattern shown in Example 16 of the textbook.

Solution Steps

  1. Step 1: Apply definition (f + g)(x) = f(x) + g(x) = (x + 1) + (2x - 3) = 3x - 2

  2. Step 2: Apply definition (f - g)(x) = f(x) - g(x) = (x + 1) - (2x - 3) = -x + 4

  3. Step 3: Apply definition (f/g)(x) = f(x)/g(x) = (x + 1)/(2x - 3), with condition g(x) ≠\neq 0, i.e., x ≠\neq 3/2

PreviousNext