Question

Complete set of values of $m$, for which function $f(x)=3+mx+e^{-x}$ is always decreasing, is

• A. $[0, \infty)$
• B. $(-\infty, 0]$
• C. $[2, 5]$
• D. $[7, 17]$

Answer 1

Answer: (B)

(2)

Answer 2

To determine the values of $m$ for which the function $f(x)=3+mx+e^{-x}$ is always decreasing, we need to ensure that its first derivative, $f'(x)$, is less than or equal to zero for all $x \in \mathbb{R}$.

Step 1: Find the first derivative of the function $f(x)$. Given $f(x) = 3 + mx + e^{-x}$.
Differentiating $f(x)$ with respect to $x$:
$f'(x) = \frac{d}{dx}(3) + \frac{d}{dx}(mx) + \frac{d}{dx}(e^{-x})$
$f'(x) = 0 + m(1) + e^{-x}(-1)$
$f'(x) = m - e^{-x}$

Step 2: Set the condition for the function to be always decreasing. For $f(x)$ to be always decreasing, we must have $f'(x) \leq 0$ for all $x \in \mathbb{R}$.
So, we set up the inequality:
$m - e^{-x} \leq 0$

Step 3: Solve the inequality for $m$. Rearranging the inequality, we get:
$m \leq e^{-x}$

This inequality must hold true for all values of $x \in \mathbb{R}$.
To satisfy $m \leq e^{-x}$ for all $x$, $m$ must be less than or equal to the minimum value that $e^{-x}$ can take.
Let's analyze the behavior of the function $g(x) = e^{-x}$:

• As $x \to \infty$, $e^{-x} \to 0$.
• As $x \to -\infty$, $e^{-x} \to \infty$.

The range of $e^{-x}$ is $(0, \infty)$. This means $e^{-x}$ is always positive and can take any value greater than 0. However, it never actually reaches 0. The infimum (greatest lower bound) of the set of values $e^{-x}$ can take is 0.

For $m \leq e^{-x}$ to hold for all $x$, $m$ must be less than or equal to this infimum.
Therefore, $m \leq 0$.

Step 4: Express the set of values of $m$. The complete set of values of $m$ for which $f(x)$ is always decreasing is $m \in (-\infty, 0]$.