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Definition of Functions

If function ...

Question

If function $f (x) = \frac{1}{2} - tan (\frac{π x}{2}); (- 1 < x < 1)$ and $g (x) = \sqrt{3 + 4 x - 4 x^{2}},$ then the domain of $(g o f)$ is

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Solution

$\begin{matrix} g (x) = \sqrt{3 + 4 x - 4 x^{2}} - (4 x^{2} - 4 x - 3) \geq 0 4 x^{2} - 4 x - 3 \leq 0 4 x^{2} - 6 x + 2 x - 3 \leq 0 (2 x + 1) (2 x - 3) \leq 0 x \in [- \frac{1}{2}, \frac{3}{2}] - \frac{1}{2} \leq \frac{1}{2} - tan (\frac{π - x}{2}) \leq \frac{3}{2} - 1 \leq tan (\frac{π x}{2}) \leq 1 - \frac{π}{4} \leq \frac{π x}{2} \leq \frac{π}{4} - \frac{1}{4} \leq \frac{x}{2} \leq \frac{1}{4} - \frac{1}{2} \leq x \leq \frac{1}{2} o m a i n o f g (f (x)) i s [- \frac{1}{2}, \frac{1}{2}] . \end{matrix}$

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Similar questions

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

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

. Find domain of

(f + g) (x)

f (x) = \sqrt{x + 3} and g (x) = x^{2} + 1

be two real functions, then find fog and gof.

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

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

Find domain of (f + g).

g (x) = 2 x + 1

f \circ g (x) = 4 x^{2} + 4 x + 2

, then range of

f (x)

f (x) = x + 5

g (x) = \sqrt{x^{2} - 9}

, then find the domain of gof (x)

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