The function f(x) =(9-x2)2 increases in

Question

The function f(x) =(9-x2)2 increases in (A)  (-3,0)∪ (3,∞ )  (B)  (-∞ ,-3)∪ (3,∞ )  (C)  (-∞ ,-3)∪ (0,3)  (D)  (-3,3)

Tardigrade Admin · Accepted Answer

Given, f(x)=(9-x2)2 On differentiating w.r.t. x, we get f(x)=2(9-x2)(-2x) Now, put f(x)=0 ⇒ 2(9-x2)(-2x)=0 ⇒ x=0,± 3 ∴ f(x) is increasing in (-3,0)∪ (3,∞ ) .

Q. The function $f (x) = (9 - x^{2})^{2}$ increases in

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

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

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

$(- 3, 3)$

$(3, \infty)$

Given, $f (x) = (9 - x^{2})^{2}$
On differentiating w.r.t. $x,$ we get $f (x) = 2 (9 - x^{2}) (- 2 x)$
Now, put $f (x) = 0$
$\Rightarrow$ $2 (9 - x^{2}) (- 2 x) = 0$
$\Rightarrow$ $x = 0, \pm 3$

$∴$ $f (x)$ is increasing in $(- 3, 0) \cup (3, \infty)$ .

Q. The function f(x)=(9−x2)2 increases in

(−3,0)∪(3,∞)

(−∞,−3)∪(3,∞)

(−∞,−3)∪(0,3)

(−3,3)

(3,∞)

Given, f(x)=(9−x2)2 On differentiating w.r.t. x, we get f(x)=2(9−x2)(−2x) Now, put f(x)=0 ⇒ 2(9−x2)(−2x)=0 ⇒ x=0,±3 ∴ f(x) is increasing in (−3,0)∪(3,∞) .