Question 30 of 35intermediate🔧 ApplyShort Answer4 marks

Let f,g:R→Rf, g : \mathbb{R} \rightarrow \mathbb{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

(f - g) (f−g)(x)=−x+4(f - g)(x) = -x + 4

(f + g) (f+g)(x)=3x−2(f + g)(x) = 3x - 2

(f/g) (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 of addition, subtraction, and quotient of real functions. For the quotient function, it is essential to state the condition that the denominator cannot equal zero, following the pattern shown in Examples 16 and 17.

Solution Steps

  1. Step 1: Addition of functions - Using (f+g)(x)=f(x)+g(x)(f + g)(x) = f(x) + g(x), we get (f+g)(x)=(x+1)+(2x−3)=3x−2(f + g)(x) = (x + 1) + (2x - 3) = 3x - 2

  2. Step 2: Subtraction of functions - Using (f−g)(x)=f(x)−g(x)(f - g)(x) = f(x) - g(x), we get (f−g)(x)=(x+1)−(2x−3)=x+1−2x+3=−x+4(f - g)(x) = (x + 1) - (2x - 3) = x + 1 - 2x + 3 = -x + 4

  3. Step 3: Quotient of functions - Using (fg)(x)=f(x)g(x)\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)}, we get (fg)(x)=x+12x−3\left(\frac{f}{g}\right)(x) = \frac{x + 1}{2x - 3}

  4. Step 4: Domain restriction - Since g(x)≠0g(x) \neq 0, we solve 2x−3=02x - 3 = 0 to get x≠32x \neq \frac{3}{2}

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