If f - g - h are real functions defined by f x - x -1- g x -1- x and h x -2 x 2-3- then find the values of 2 f - g - h 1 and 2 f - g - h 0-

We have f(x) = √(x+1), g(x) = 1/x and h(x) = 2x^2 3 Clearly, f(x) is defined for x+1 ≥ 0 ⇒ x ≥ 1 ⇒ x ϵ [ 1, ∞] g(x) is defined for x ≠ 0 ⇒ x ϵ R {0} and h( ...

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Standard XII

Mathematics

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If f , g , h ...

Question

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

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Solution

We have
$f (x) = \sqrt{x + 1}, g (x) = \frac{1}{x}$
and $h (x) = 2 x^{2} - 3$
Clearly, f(x) is defined for $x + 1 \geq 0$
$\Rightarrow x \geq - 1 \Rightarrow x ϵ [- 1, \infty]$
g(x) is defined for $x \neq 0$
$\Rightarrow x ϵ R - {0}$
and h(x) is defined for all $x ϵ R$
$∴ Domain (f) \cap Domain (g) \cap Domain (h) = [- 1, \infty] - {0}$
Clearly,
$2 f + g - h : [- 1, \infty] - {0} \to R$ is given by
$(2 f + g - h) (x) = 2 f (x) + g (x) - h (x) = 2 \sqrt{x + 1} + \frac{1}{x} - 2 x^{2} + 3 ∴ (2 f + g - h) (1) = 2 \sqrt{1 + 1} + \frac{1}{1} - 2 \times (1)^{2} + 3 = 2 \sqrt{2} + 1 - 2 + 3 = 2 \sqrt{2} + 4 - 2 = 2 \sqrt{2} + 2$
and, (2f+g-h)(0) does not exist, it is not lies in the domain $x ϵ [- 1, \infty] - {0}$ .

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If f, g, h are real functions defined by f(x)=√x+1,g(x)=1x and h(x)=2x2−3, then find the values of (2f + g - h) (1) and (2f + g - h) (0).

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