Let f:R o R be defined as f(x) = \left\{ {\begin{matrix} { - 55x,} & {if\,x 4,} \ \end{matrix} } ight. Let A = {x \in R : f is increasing}. Then A is equal to :

Question

Let f:R 	o R be defined as f(x) = \left\{ {\begin{matrix} { - 55x,} & {if\,x < - 5} \ {2{x^3} - 3{x^2} - 120x,} & {if\, - 5 \le x \le 4} \ {2{x^3} - 3{x^2} - 36x - 336,} & {if\,x > 4,} \ \end{matrix} } ight. Let A = {x \in R : f is increasing}. Then A is equal to :

Accepted Answer

Answer: C. ( - 5, - 4) \cup (4,\infty ) f(x) = \left\{ {\begin{matrix} { - 55x,} & {if\,x < - 5} \ {2{x^3} - 3{x^2} - 120x,} & {if\, - 5 \le x \le 4} \ {2{x^3} - 3{x^2} - 36x - 336,} & {if\,x > 4,} \ \end{matrix} } ight. Now, f'(x) = \left\{ {\begin{matrix} { - 55} & ; & {x < - 5} \ {6({x^2} - x - 20)} & ; & { - 5 < x < 4} \ {6({x^2} - x - 6)} & ; & {x > 4} \ \end{matrix} } ight. f'(x) = \left\{ {\begin{matrix} { - 55} & ; & {x < - 5} \ {6(x - 5)(x + 4)} & ; & { - 5 < x < 4} \ {6(x - 3)(x + 2)} & ; & {x > 4} \ \end{matrix} } ight. Hence, f(x) is monotonically increasing in interval ( - 5, - 4) \cup (4,\infty )

Let $$f:R \to R$$ be defined as $$f(x) = \left\{ {\begin{matrix} { - 55x,} & {if\,x < - 5} \\ {2{x^3} - 3{x^2} - 120x,} & {if\, - 5 \le x \le 4} \\ {2{x^3} - 3{x^2} - 36x - 336,} & {if\,x > 4,} \\ \end{matrix} } \right.$$ Let A = {x $$ \in $$ R : f is increasing}. Then A is equal to :

Answer

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