Question

If $f\left(x\right)=e^{-x}$, then $\frac{f\left(-a\right)}{f\left(b\right)}$ is equal to

• A. $f\left(a+b\right)$
• B. $f\left(a-b\right)$
• C. $f\left(-a+b\right)$
• D. $f\left(-a-b\right)$

Answer 1

The correct option is D

$f\left(-a-b\right)$

Explanation for the correct option.

Find the value of the given expression.

For the function $f\left(x\right)=e^{-x}$, the value of $\frac{f\left(-a\right)}{f\left(b\right)}$ is given as:

$\begin{array}{rcl}\frac{f\left(-a\right)}{f\left(b\right)}& =& \frac{e^{-\left(-a\right)}}{e^{-b}}\\ & =& \frac{e^a}{e^{-b}}\\ & =& e^{a-\left(-b\right)}\left[\frac{a^m}{a^n}=a^{m-n}\right]\\ & =& e^{a+b}\\ & =& e^{-\left(-a-b\right)}\\ & =& f\left(-a-b\right)\left[f\left(x\right)=e^{-x}\right]\end{array}$

Hence, the correct option is D.