question_answer

Find the intervals in which the function                                                       \[f(x)=2{{x}^{3}}-15{{x}^{2}}+36x+17\]  is increasing or decreasing.

Answer:

The given function is                                                       \[f(x)=2{{x}^{3}}-15{{x}^{2}}+36x+17\]             On differentiating both sides w.r.t. x, we get                                                                   \[f'(x)=6{{x}^{2}}-30x+36\] Now, put                                                                                  \[f'(x)=0\]                                                                               \[\Rightarrow\]                                                                    \[6{{x}^{2}}-30x+36=0\] .                                                                                                \[6({{x}^{2}}-5x+6)=0\] \[\Rightarrow \]   \[6(x-2)(x+3)=0\] \[\Rightarrow \]   x = 2, 3 These two values divides the number line into three disjoint sub-intervals \[(-\infty ,\,2),\] (2, 3) and \[(3,\,\infty ).\] The sign of \[f'(x)\] in each interval is given below

interval	\[\mathbf{f'(x)=6(x-2)(x-3)}\]	Sign of \[\mathbf{f'(x)}\]
\[(-\,\infty ,\,\,2)\]	\[(+)(-)(-)\]	\[\text{+ ve}\]
(2, 3)	\[(+)(+)(-)\]	\[-ve\]
\[(3,\,\,\infty )\]	\[(+)(+)(+)\]	+ ve

We know that, f(x} is said to be an increasing function \[f'(x)\ge 0\] and decreasing function when \[f'(x)\le 0.\] So, f (x) is increasing on \[(-\,\infty ,\,\,2)\] or \[(3,\,\,\infty )\] and it is decreasing on (2, 3). Now, as f(x) being a polynomial function, so it continuous at x = 2, 3. \[\therefore \] Given function is increasing in the intervals \[\left( -\,\infty ,\,\,2 \right]\] or \[\left[ 3,\,\,\infty  \right),\] and decreasing in the interval [2, 3].