Prove that the function f(x) = 4x ³ + 6x ² – 24x + 1 decreases in the interval (–2, 1)

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$$\begin{gathered} \mathrm{We} \mathrm{have} \\ \mathrm{f}\left(\mathrm{x}\right)=4{\mathrm{x}}^3+6{\mathrm{x}}^2-24\mathrm{x}+1 \\ \mathrm{Differentiate} \mathrm{with} \mathrm{respect} \mathrm{to} \mathrm{x} \\ \mathrm{f}\text{'}\left(\mathrm{x}\right)=12{\mathrm{x}}^2+12\mathrm{x}-24=12\left({\mathrm{x}}^2+\mathrm{x}-2\right) \\ =12\left(\mathrm{x}-1\right)\left(\mathrm{x}+2\right) \\ \mathrm{Using} \mathrm{method} \mathrm{of} \mathrm{interval} \\ -\infty +++++-2----1+++++\infty \\ \mathrm{As} \mathrm{f}\text{'}\left(\mathrm{x}\right)<0 \mathrm{for} \mathrm{x}\in \left(-2,1\right) \\ \mathrm{Hence} \mathrm{f}\left(\mathrm{x}\right) \mathrm{decreases} \mathrm{for} \mathrm{x}\in \left(-2,1\right) \end{gathered}$$

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