Question

Let \(f(x)=\left(2x^{2}-9x+11\right)\cdot e^{x},\ -\infty<x<\infty\). Then \(f\) has :

• A. a local maxima at \(x=\frac{1}{2}\) and a local minima at \(x=2\).
• B. a local minima at \(x=\frac{1}{2}\) and a local maxima at \(x=2\).
• C. neither a local maxima nor a local minima at \(x=\frac{1}{2}\).
• D. neither a local maxima nor a local minima at \(x=2\).

Hint:

Differentiate using the product rule and study the sign change of \(f'(x)\) at its zeros.

Answer 1

The Correct Option is A

\(f'(x)=e^{x}\left(2x^{2}-9x+11\right)+e^{x}(4x-9)=e^{x}\left(2x^{2}-5x+2\right)=e^{x}(2x-1)(x-2)\).
So \(f'(x)=0\) at \(x=\frac{1}{2}\) and \(x=2\).
Since \(e^{x}>0\), the sign of \(f'\) is the sign of \((2x-1)(x-2)\): positive for \(x<\frac{1}{2}\), negative for \(\frac{1}{2}<x<2\), positive for \(x>2\).
At \(x=\frac{1}{2}\), \(f'\) changes from + to -, so there is a local maximum. At \(x=2\), \(f'\) changes from - to +, so there is a local minimum.
Answer: (1) a local maxima at \(x=\frac{1}{2}\) and a local minima at \(x=2\).