Let \( f: R \rightarrow R \) be defined as\[f(x)=\left\{\begin{array}{

1.	Let \( f: R \rightarrow R \) be defined as\[f(x)=\left\{\begin{array}{ll}-55 x, & \text { if } x4\end{array}\right.\]Let \( A =\{x \in R : f \) is increasing \( \} . \) Then \( A \) is equal to :
Answer» f(x) = \(\begin{cases}-55x&,x<-5\\ 2x^3-3x^2-120x&,-5\leq x\leq4\\2x^3&,x>4\end{cases}\) f'(x) = \(\begin{cases}-55&x<-5\\6x^2-6x-120&,-5<4<4\\6x^2-6x-36&,x>4\end{cases}\) For f'(x) \(\geq\) 0 Case I :- -5 < x < 4 Then f'(x) \(\geq0\) implies 6x² - 6x - 120 \(\geq\) 0 ⇒ x² - x - 20 \(\geq\) 0 ⇒ (x - 5) (x + 4) \(\geq\) 0 ⇒ x \(\leq\) - 4 or x \(\geq\) 5 But -5 <x < 4 \(\therefore\) -5 < x < 4 Hence, function f(x) increases in domain x \(\in\)(-5, -4) Case II: x > 4 Then f'(x) \(\geq\) 0 implies 6x² - 6x - 36 \(\geq\) 0 ⇒ x² - x - 6 \(\geq\) 0 ⇒ (x - 3) (x + 2) \(\geq\) 0 ⇒ x \(\leq\) - 2 or x \(\geq\) 3 But x > 4 Hence, f(x) increases in (4, \(\infty\)) From both cases, we can say that f(x) is increases in (-5, -4) U (4, \(\infty\)). A = {x\|x \(\in\) (-5, -4) U (4, \(\infty\))}.

Let \( f: R \rightarrow R \) be defined as\[f(x)=\left\{\begin{array}{ll}-55 x, & \text { if } x4\end{array}\right.\]Let \( A =\{x \in R : f \) is increasing \( \} . \) Then \( A \) is equal to :

Answer»

f(x) = \(\begin{cases}-55x&,x<-5\\ 2x^3-3x^2-120x&,-5\leq x\leq4\\2x^3&,x>4\end{cases}\)

f'(x) = \(\begin{cases}-55&x<-5\\6x^2-6x-120&,-5<4<4\\6x^2-6x-36&,x>4\end{cases}\)

For f'(x) \(\geq\) 0

Case I :- -5 < x < 4

Then f'(x) \(\geq0\) implies

6x² - 6x - 120 \(\geq\) 0

⇒ x² - x - 20 \(\geq\) 0

⇒ (x - 5) (x + 4) \(\geq\) 0

⇒ x \(\leq\) - 4 or x \(\geq\) 5

But -5 <x < 4

\(\therefore\) -5 < x < 4

Hence, function f(x) increases in domain x \(\in\)(-5, -4)

Case II: x > 4

Then f'(x) \(\geq\) 0 implies

6x² - 6x - 36 \(\geq\) 0

⇒ x² - x - 6 \(\geq\) 0

⇒ (x - 3) (x + 2) \(\geq\) 0

⇒ x \(\leq\) - 2 or x \(\geq\) 3

But x > 4

Hence, f(x) increases in (4, \(\infty\))

From both cases, we can say that f(x) is increases in (-5, -4) U (4, \(\infty\)).

A = {x|x \(\in\) (-5, -4) U (4, \(\infty\))}.