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Worked Examples · Example 16

Q.Let f(x)=x2f(x) = x^2 and g(x)=2x+1g(x) = 2x + 1 be two real functions. Find (f+g)(x)(f + g)(x), (f−g)(x)(f - g)(x), (fg)(x)(fg)(x), (fg)(x)\left(\dfrac{f}{g}\right)(x).

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Function operations combine two functions pointwise: add, subtract, multiply, or divide their outputs for the same input. For f(x)=x2f(x)=x^2 and g(x)=2x+1g(x)=2x+1, we get (f+g)(x)=x2+2x+1(f+g)(x)=x^2+2x+1, (f−g)(x)=x2−2x−1(f-g)(x)=x^2-2x-1, (fg)(x)=2x3+x2(fg)(x)=2x^3+x^2, and (fg)(x)=x22x+1\left(\frac{f}{g}\right)(x)=\frac{x^2}{2x+1} (with x≠−12x\neq -\frac12).

When you have two functions ff and gg, you can combine them using ordinary arithmetic — but applied pointwise. That means for each xx in the domain, you evaluate f(x)f(x) and g(x)g(x) separately, then perform the operation. The result is a new function whose rule is that combination.

This is exactly like adding or multiplying numbers, except the numbers come from plugging xx into each function. The only catch is division: you must exclude any xx that makes the denominator zero, because division by zero is undefined.

Let’s work through each operation step by step.


  1. Addition: (f+g)(x)(f+g)(x) By definition, (f+g)(x)=f(x)+g(x)(f+g)(x) = f(x) + g(x). So:

(f+g)(x)=x2+(2x+1)=x2+2x+1.(f+g)(x) = x^2 + (2x + 1) = x^2 + 2x + 1.

Notice that x2+2x+1x^2 + 2x + 1 factors as (x+1)2(x+1)^2, but the simplified form is perfectly fine.

  1. Subtraction: (f−g)(x)(f-g)(x) Here (f−g)(x)=f(x)−g(x)(f-g)(x) = f(x) - g(x).

(f−g)(x)=x2−(2x+1)=x2−2x−1.(f-g)(x) = x^2 - (2x + 1) = x^2 - 2x - 1.

No further simplification is needed — it’s a quadratic expression.

  1. Multiplication: (fg)(x)(fg)(x) The product is (fg)(x)=f(x)⋅g(x)(fg)(x) = f(x) \cdot g(x).

(fg)(x)=x2⋅(2x+1)=2x3+x2.(fg)(x) = x^2 \cdot (2x + 1) = 2x^3 + x^2.

Just distribute x2x^2 across the binomial.

  1. Division: (fg)(x)\left(\frac{f}{g}\right)(x) For division, (fg)(x)=f(x)g(x)\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)}, provided g(x)≠0g(x) \neq 0.

(fg)(x)=x22x+1.\left(\frac{f}{g}\right)(x) = \frac{x^2}{2x + 1}.

Now, g(x)=2x+1=0g(x) = 2x + 1 = 0 when x=−12x = -\frac12. So the domain of this new function is all real numbers except −12-\frac12.

Watch out

A common mistake is to forget the domain restriction for division. The expression x22x+1\frac{x^2}{2x+1} is not defined at x=−12x = -\frac12, so you must explicitly state that x≠−12x \neq -\frac12 when writing the quotient function.

Tip

Notice that (f+g)(x)=x2+2x+1=(x+1)2(f+g)(x) = x^2 + 2x + 1 = (x+1)^2. This is a perfect square — a neat observation, but not required for the answer. It can help you check your work quickly.

✓Final answer

The results are (f+g)(x)=x2+2x+1(f+g)(x) = x^2 + 2x + 1, (f−g)(x)=x2−2x−1(f-g)(x) = x^2 - 2x - 1, (fg)(x)=2x3+x2(fg)(x) = 2x^3 + x^2, and (fg)(x)=x22x+1\left(\frac{f}{g}\right)(x) = \frac{x^2}{2x+1} for x≠−12x \neq -\frac12.

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