If f, g and h are real functions defined by f(x)

Question:

If $t, g$ and $h$ are real functions defined by $f(x)=\sqrt{x+1}, g(x)=\frac{1}{x}$ and $h(x)=2 x^{2}-3$, find the values of $(2 f+g-h)(1)$ and $(2 f+g-h)(0)$.

Solution:

Given:

$f(x)=\sqrt{x+1}, g(x)=\frac{1}{x}$ and $h(\mathrm{x})=2 \mathrm{x}^{3}-3$

Clearly, f (x) is defined for x + 1 ≥ 0 .

$\Rightarrow x \geq-1$

$\Rightarrow x \in[-1, \infty]$

Thus, domain ( f ) = [ -">-1, ∞] .

Clearly, g (x) is defined for x ≠ 0 .

⇒ x ∈ R – { 0} and h(x) is defined for all x such that  x ∈ R .

Thus,

domain ( ) ∩ domain (g) ∩ domain (h) = [ -">- 1, ∞] – { 0}.

Hence,

$(2 f+g-h):[-1, \infty]-\{0\} \rightarrow \mathrm{R}$ is given by:

$(2 f+g-h)(x)=2 f(x)+g(x)-h(x)$

$=2 \sqrt{x+1}+\frac{1}{x}-2 x^{2}+3$

$(2 f+g-h)(1)=2 \sqrt{2}+1-2+3=2 \sqrt{2}+4-2=2 \sqrt{2}+2$

$(2 f+g-h)(0)$ does not exist because 0 does not lie in the domain $x \in[-1, \infty]-\{0\}$