The interval in which the function \( f(x)=x^{3}-6 x^{2}+9 x+10 \) is increasing in

• Correct answer
  A

  \((-\infty, 1] \cup[3, \infty) \)
• B

  \((-\infty,-1] \cup[3, \infty) \)
• C

  \( (-\infty, 1) \cup(3, \infty) \)
• D

  \( [1,3] \)

Solution

\(\begin{aligned} &f(x)=x^{3}-6 x^{2}+9 x+10 \\ &f^{\prime}(x)=3 x^{2}-12 x+9 \\ &f^{\prime}(x)=3\left(x^{2}-4 x+3\right) \\ &f^{\prime}(x)=3(x-1)(x-3) \end{aligned}\) For \(f(x)\) to be inereasing, we must have \(\begin{aligned} & F^{\prime}(x) > 0 \\ \Rightarrow & 3(x-1)(x-3) > 0 \\ \Rightarrow &(x-1)(x-3) > 0 \\ \Rightarrow & x < 1 \text { or } x > 3 \\ \Rightarrow & x \in(-\infty, 1) \cup(3, \infty) \\ & \text { SO, } F(x) \text { is increasing on } x \in(-\infty, 1) \cup(3, \infty) . \end{aligned}\) For \(f(x)\) to be decrecsing, we must hare \(F^{\prime}(x) < 0\) \(\begin{aligned} &\Rightarrow 3(x-1)(x-3) < 0 \\ &\Rightarrow(x-1)(x-3) < 0 \\ &\Rightarrow 1 < x < 3 \\ &\Rightarrow x \in(1,3) \end{aligned}\) so, \(f(x)\) is decreasing on \(x \in(1,3)\).

Hence, the correct answer is option: 'A'.