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Maths-

General

Easy

Question

The function $f open parentheses x close parentheses equals open parentheses 9 minus x to the power of 2 end exponent close parentheses to the power of 2 end exponent$ increasing in

$(- 3, 0) \cup (3, \infty)$
$(- \infty, - 3) \cup (3, \infty)$
$(- \infty, - 3) \cup (0, 3)$
$(- 3, 3)$

The correct answer is: $open parentheses negative 3 comma blank 0 close parentheses union open parentheses 3 comma blank infinity close parentheses$

Given, $f open parentheses x close parentheses equals open parentheses 9 minus x to the power of 2 end exponent close parentheses to the power of 2 end exponent$
$rightwards double arrow blank f to the power of ´ ´ end exponent open parentheses x close parentheses equals 2 open parentheses 9 minus x to the power of 2 end exponent close parentheses left parenthesis negative 2 x right parenthesis$

Now, put $f to the power of ´ ´ end exponent open parentheses x close parentheses equals 0$

$rightwards double arrow blank 2 open parentheses 9 minus x to the power of 2 end exponent close parentheses open parentheses negative 2 x close parentheses equals 0 blank rightwards double arrow blank x equals 0 comma blank plus-or-minus 3$

$therefore blank f to the power of ´ ´ end exponent left parenthesis x right parenthesis$ is increasing in $left parenthesis negative 3 comma blank 0 right parenthesis union left parenthesis 3 comma blank infinity right parenthesis$

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