Question:
The function \(f(x)=\mathrm{e}^{x+1}\left(4 x^{2}-16 x+11\right)\) is :
The function \(f(x)=\mathrm{e}^{x+1}\left(4 x^{2}-16 x+11\right)\) is :
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Differentiate and factorise the quadratic factor of \(f'(x)\); the sign of \(f'\) decides where \(f\) increases or decreases.
Updated On: Nov 12, 2026
- decreasing in \(\left(-\infty, \frac{1}{2}\right) \cup\left(\frac{5}{2}, \infty\right)\)
- increasing in \((-\infty,-2) \cup(2, \infty)\)
- decreasing in \(\left(-\infty,-\frac{5}{2}\right) \cup\left(\frac{1}{2}, \infty\right)\)
- increasing in \(\left(-\infty,-\frac{1}{2}\right) \cup\left(\frac{5}{2}, \infty\right)\)
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The Correct Option is D
Solution and Explanation
\(f'(x)=e^{x+1}\left(4x^2-16x+11\right)+e^{x+1}(8x-16)=e^{x+1}\left(4x^2-8x-5\right)\).
\(4x^2-8x-5=(2x-5)(2x+1)\), with roots \(x=\frac52\) and \(x=-\frac12\).
Since \(e^{x+1}>0\), \(f'(x)>0\) when \(x<-\frac12\) or \(x>\frac52\), and \(f'(x)<0\) for \(-\frac12<x<\frac52\).
So \(f\) is increasing in \(\left(-\infty,-\frac12\right)\cup\left(\frac52,\infty\right)\).
Answer: (4) increasing in \(\left(-\infty,-\frac{1}{2}\right) \cup\left(\frac{5}{2}, \infty\right)\)
\(4x^2-8x-5=(2x-5)(2x+1)\), with roots \(x=\frac52\) and \(x=-\frac12\).
Since \(e^{x+1}>0\), \(f'(x)>0\) when \(x<-\frac12\) or \(x>\frac52\), and \(f'(x)<0\) for \(-\frac12<x<\frac52\).
So \(f\) is increasing in \(\left(-\infty,-\frac12\right)\cup\left(\frac52,\infty\right)\).
Answer: (4) increasing in \(\left(-\infty,-\frac{1}{2}\right) \cup\left(\frac{5}{2}, \infty\right)\)
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