Fx-1-2-tan - x-2 -1 - x - 1-and gx -3-4x-4x2Find domain of f - g-

4x^2 4x 3 ≤ 0 (2x+1)(2x 3)≤ 0 ⇒ x ∈[ 1/2 , nbsp;3/2], x ∈( 1,1) ∴ x ∈[ 1/2 , 1 ) ...

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Fractional Part Function

fx=1/2-tan π ...

Question

$f (x) = \frac{1}{2} - t a n (\frac{π x}{2}) [- 1 < x < 1]$

and g(x) = $\sqrt{(3 + 4 x - 4 x^{2})}$

Find domain of (f + g).

$[\frac{- 1}{2}, 1)$

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$(\frac{- 1}{2}, 1]$

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$[- \frac{1}{2}, \frac{3}{2}]$

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$(- 1, 1)$

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Solution

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Similar questions

$f (x) = \frac{1}{2} - t a n (\frac{π x}{2}) [- 1 < x < 1]$

and g(x) = $\sqrt{(3 + 4 x - 4 x^{2})}$

Find domain of (f + g).

f (x) = \frac{1}{2} - tan (\frac{π x}{2}) - 1 < x < 1

g (x) = \sqrt{(3 + 4 x - 4 x^{2})}

(f + g) (x)

f (x) = \frac{1}{2} - tan (\frac{π x}{2}), - 1 < x < 1

g (x) = \sqrt{(3 + 4 x - 4 x^{2})}

(f + g)

f (x) = \frac{1}{2} - tan (\frac{π x}{2}); (- 1 < x < 1)

g (x) = \sqrt{3 + 4 x - 4 x^{2}},

(g o f)

f (x) = e^{x}, g (x) = \log_{e} x

f (x) = x^{2}, g (x) = \cos x

f (x) = | x |, g (x) = \sin x

f (x) = x + 1, g (x) = e^{x}

f (x) = \sin^{- 1} x, g (x) = x^{2}

f (x) = x + 1, g (x) = \sin x

f (x) = x + 1, g (x) = 2 x + 3

f (x) = c, c \in R, g (x) = \sin x^{2}

f (x) = x^{2} + 2, g (x) = 1 - \frac{1}{1 - x}

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