Let $f (x) = \sqrt{x}$ and $g (x) = x$ be two functions defined over the set of non-negative real numbers. Find:

(i) $(f + g) (x)$

(ii) $(f - g) (x)$

(iii) $(f g) (x)$

(iv) $\frac{f}{g} (x)$

Open in App

Solution

Here $f : [0, \infty) \to R : f (x) = \sqrt{x}$ and $g : [0, \infty) \to R : g (x) = x$

$∴ dom (f) = [0, \infty)$ and $dom (g) = [0, \infty)$

So, $dom (f) \cap dom (g) = [0, \infty) \cap [0, \infty) = [0, \infty)$

(i) $(f + g) : [0, \infty) \to R$ is given by

$(f + g) (x) = f (x) + g (x) = (\sqrt{x} + x)$

(ii) $(f - g) : [0, \infty) \to R$ is given by

$(f - g) (x) = f (x) - g (x) = (\sqrt{x} - x)$

(iii) $(f g) : [0, \infty) \to R$ is given by

$(f g) (x) = f (x) . g (x) = (\sqrt{x} \times x) = x^{\frac{3}{2}}$

(iv) ${x : g (x) = 0} = {0}$

$∴ dom (\frac{f}{g}) = dom (f) \cap dom (g) - {x : g (x) = 0}$

$= [0, \infty) \cap [0, \infty) - {0} = (0, \infty)$

So, $\frac{f}{g} : (0, \infty) \to R$ is given by

$(\frac{f}{g}) (x) = \frac{f (x)}{g (x)} = \frac{\sqrt{x}}{x} = \frac{1}{\sqrt{x}}, x \neq 0$ .

Suggest Corrections

Similar questions

Let f and g be real functions, defined by $f (x) = \sqrt{x + 2}$ and $g (x) = \sqrt{4 - x^{2}}$ . Find

(i) $(f + g) (x)$

(ii) $(f - g) (x)$

(iii) $(f g) (x)$

(iv) $(f f) (x)$

(v) $(g g) (x)$

(vi) $(\frac{f}{g}) (x)$

f (x) = \sqrt{x}

g (x) = x

(f + g) (f g)

(f / g)

Let $f : R \to R : f (x) = x^{2}$ and $g : R \to R : g (x) = 2 x + 1$ . Find

(i) $(f + g) (x)$

(ii) $(f - g) (x)$

(iii) $(f g) (x)$

(iv) $(\frac{f}{g}) (x)$

f (x) = x^{2}

g (x) = 2 x + 1

(f + g) (x), (f - g) (x), (f g) (x), (\frac{f}{g}) (x) .

Let $f (x) = x^{2}$ and $g (x) = 2 x + 1$ be two real functions. Find (f + g), (f - g) (x), (fg) (x) and $(\frac{f}{g}) (x)$

Join BYJU'S Learning Program