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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).

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Solution

## 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) $=2\sqrt{x+1}+\frac{1}{x}-2{x}^{2}+3\phantom{\rule{0ex}{0ex}}$ $\left(2f+g-h\right)\left(1\right)=2\sqrt{2}+1-2+3=2\sqrt{2}+4-2=2\sqrt{2}+2$ (2f + g – h) (0) does not exist because 0 does not lie in the domain x ∈[ - 1, ∞] – {0}.

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