Let $f:R \to R$ be defined as
$$f(x) = \left\{ {\matrix{
{ - 55x,} & {if\,x < - 5} \cr
{2{x^3} - 3{x^2} - 120x,} & {if\, - 5 \le x \le 4} \cr
{2{x^3} - 3{x^2} - 36x - 336,} & {if\,x > 4,} \cr
} } \right.$$
Let A = {x $\in$ R : f is increasing}. Then A is equal to :
Solution
$$f(x) = \left\{ {\matrix{
{ - 55x,} & {if\,x < - 5} \cr
{2{x^3} - 3{x^2} - 120x,} & {if\, - 5 \le x \le 4} \cr
{2{x^3} - 3{x^2} - 36x - 336,} & {if\,x > 4,} \cr
} } \right.$$
<br><br>Now, $$f'(x) = \left\{ {\matrix{
{ - 55} & ; & {x < - 5} \cr
{6({x^2} - x - 20)} & ; & { - 5 < x < 4} \cr
{6({x^2} - x - 6)} & ; & {x > 4} \cr
} } \right.$$<br><br>$$f'(x) = \left\{ {\matrix{
{ - 55} & ; & {x < - 5} \cr
{6(x - 5)(x + 4)} & ; & { - 5 < x < 4} \cr
{6(x - 3)(x + 2)} & ; & {x > 4} \cr
} } \right.$$<br><br>Hence, f(x) is monotonically increasing in interval $( - 5, - 4) \cup (4,\infty )$
About this question
Subject: Mathematics · Chapter: Application of Derivatives · Topic: Tangents and Normals
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