The function $\mathrm{f}(x)=2 x^3-6 x+5$ is an increasing function, if

The function $\mathrm{f}(x)=2 x^3-6 x+5$ is an increasing function, if
  1. $0 \lt x \lt 1$
  2. $-1 \lt x \lt 1$
  3. $x \lt -1$ or $x\gt1$
  4. $-1 \lt x \lt -\frac{1}{2}$

Solution

$\begin{aligned} & f(x)=2 x^3-6 x+5 \\ \therefore \quad & f^{\prime}(x)=6 x^2-6 \end{aligned}$
For $\mathrm{f}(x)$ to be increasing, $\mathrm{f}^{\prime}(x)\gt0$ $\begin{aligned} & \Rightarrow 6 x^2-6\gt0 \Rightarrow(x-1)(x+1)\gt0 \\ & \Rightarrow x\gt1 \text { or } x \lt -1 \end{aligned}$

Asked in: MHT CET 2024 (09 May Shift 2)

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