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