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Find the range of the following functions given by

i) $f\left ( x \right )= \frac{3}{2-x^2}$

ii) $f\left ( x \right )= 1-\left | x-2 \right |$

iii) $f\left ( x \right )= \left | x-3 \right |$

iv) $f\left ( x \right )= 1+3\cos 2x$

Answers (1)

i)Given data: $f\left ( x \right )= \frac{3}{2-x^2}$

Let us consider that, y = f(x)

Thus, $y = 3/(2 - x^2)$

Thus,

$y(2 - x^2) = 3$

$2y - yx^2 = 3$

$yx^2 = 2y - 3$

$x^2 = 2 - 3/y$

x is real, if $2y - 3 \geq0 \: \: and\: \: y \geq0$

Thus, $y \geq 3/2$

Therefore,

Range of $f = (3/2, \infty)$

ii)Given data: $f\left ( x \right )= 1-\left | x-2 \right |$

Now, we know that,

$\left |x-2 \right | = -(x-2) , if x < 2, \: \: and\: \: \left | x-2 \right |= (x-2), if \geq 2$

Thus, $\left |-x-2 \right | \geq 0$

Thus, $1 -\left | x-2 \right | \leq 1$

Therefore,

Range of $f = (-\infty, 1)$

iii)Given data: $f\left ( x \right )= \left | x-3 \right |$

Now, we know that,

$\left |x-3 \right | \geq 0$

Thus, $f(x) = 0$

Therefore, range of $f = (0,\infty)$

iv)Given data: $f\left ( x \right )= 1+3\cos 2x$

Now, we know that,

$-1 \leq cos 2x \leq 1$

Thus, $-3 \leq 3 cos 2x \leq 3$

$-3 + 1 \leq 1 + 3 cos 2x \leq 3 + 1$

Thus, $-2 \leq 1 + 3 cos 3x \leq 4$

$-2 \leq f(x) \leq 4$

Therefore, range of $f = [-2, 4].$

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infoexpert21

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