Question

Let f(x) = 2x + 5 and g(x) = x² + x. Describe (i) f + g (ii) f − g (iii) fg (iv) f/g. Find the domain in each case.

 

Answer 1

Given:
f(x) = 2x + 5 and g(x) = x² + x
Clearly, f (x) and g (x) assume real values for all x.
Hence,
domain (f) = R and domain (g) = R.

\[\therefore D\left( f \right) \cap D\left( g \right) = R\]

Now,
(i) (f + g) : R → R is given by (f + g) (x) = f (x) + g (x) = 2x + 5 + x² + x = x² + 3x + 5.
 Hence, domain ( f + g) = R .

(ii) (f - g) : R → R is given by (f - g) (x) = f (x) - g (x) = (2x + 5) - (x² + x) = 5 + x -x²
 Hence, domain ( f -g) = R.

(iii) (fg) : R → R is given by (fg) (x) = f(x).g(x) = (2x + 5)(x² + x)
                                                                   = 2x³ + 2x² + 5x² +5x
                                                                     = 2x³ + 7x² + 5x
Hence, domain ( f.g) = R .

(iv) Given:
 g(x) = x² + x
g(x) = 0 ⇒ x² + x = 0 = x(x+ 1) = 0
⇒ x = 0 or (x + 1) = 0
⇒ x = 0 or x = - 1

Now , 

\[\frac{f}{g}: R - \left\{ - 1, 0 \right\} \to R \text{ is given by } \left( \frac{f}{g} \right)\left( x \right) = \frac{f\left( x \right)}{g\left( x \right)} = \frac{2x + 5}{x^2 + x}\]

Hence,

\[\text{ domain } \left( \frac{f}{g} \right) = R - \left\{ - 1, 0 \right\}\]