Find Range of $f (x) :$

$f (x) = 2$ $s i n$ $x$ , $0$ $⩽$ $x$ $⩽$ $π$

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Solution

$f (x) = 2 sin x ϵ [0. π]$

$f (x) = 2 [0, 1]$

$f (x) = [0, 2]$

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f (x) = \frac{({sin}^{- 1} x + {tan}^{- 1} x)}{π} + 2 \sqrt{x}

f : R \to R

f (x) = \frac{s i n [x] π + t a n [x] π}{1 + [x^{2}]}

Write the range of the funciton $f (x) = s i n [x]$ , where $\frac{- π}{4} \leq x \leq \frac{π}{4}$ .

- \frac{π}{4} \leq x \leq \frac{π}{4}

f (x) = ∣ ∣ {sin}^{- 1} (sin x) ∣ ∣ - (\frac{π - x}{2})

f (x) = 0

x \in [- π, π]

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