Question

If the function $e^{-x}f\left(x\right)$ assumes its minimum in the interval $[0,1]\text{at}x=\frac{1}{4},$ which of the following is true?

• A. $f^{\prime }\left(x\right)<f\left(x\right),0<x<\frac{1}{4}$
• B. $f^{\prime }\left(x\right)<f\left(x\right),\frac{1}{4}<x<\frac{3}{4}$
• C. $f^{\prime }\left(x\right)<f\left(x\right),\frac{3}{4}<x<1$
• D. $f^{\prime }\left(x\right)>f\left(x\right),0<x<\frac{1}{4}$

Answer 1

The correct option is A $f^{\prime }\left(x\right)<f\left(x\right),0<x<\frac{1}{4}$

$\frac{d}{dx}\left(y{e}^{-x}\right)$ is an increasing function.

$\begin{array}{cc}0<x<\frac{1}{4}& x>\frac{1}{4}\\ \frac{d}{dx}\left(y{e}^{-x}\right)<0& \frac{d}{dx}\left(y{e}^{-x}\right)>0\\ e^{-x}\frac{dy}{dx}-e^{-x}y<0& e^{-x}\frac{dy}{dx}-e^{-x}y>0\\ \frac{dy}{dx}<y& \frac{dy}{dx}>y\\ f^{\prime }\left(x\right)<f\left(x\right)& f^{\prime }\left(x\right)>f\left(x\right)\end{array}$