Question

If f, g and h are real functions defined by $f\left(x\right)=\sqrt{x+1},g\left(x\right)=\frac{1}{x}$and h(x) = 2x² − 3, find the values of (2f + g − h) (1) and (2f + g − h) (0).

Answer 1

Given:
$f\left(x\right)=\sqrt{x+1},g\left(x\right)=\frac{1}{x}\mathrm{and}h\left(\mathrm{x}\right)=2{\mathrm{x}}^3-3$
Clearly, f (x) is defined for x + 1 ≥ 0 .
⇒ x ≥ $-$ 1
⇒ x ∈ [ $-$1, ∞]
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 ( f ) ∩ domain (g) ∩ domain (h) = [ $-$ 1, ∞] – { 0}.
Hence,
(2f + g – h) : [ $-$ 1, ∞] – { 0} → R is given by:
(2f + g – h)(x) = 2f (x) + g (x) $-$ h (x)
$=2\sqrt{x+1}+\frac{1}{x}-2x^2+3$
$(2f+g-h)\left(1\right)=2\sqrt{2}+1-2+3=2\sqrt{2}+4-2=2\sqrt{2}+2$
(2f + g – h) (0) does not exist because 0 does not lie in the domain x ∈[ - 1, ∞] – {0}.