The function defined by f(x) = (x + 2) e^(-x) is (A) decreasing for all x (B) decreasing in (-∞, -1)

Q: Wait, why do we reject the negative root here? Or rather, why is $e^{-x}$ never zero?

That is a fantastic question! Remember the fundamental nature of the exponential function $f(x) = a^x$ (where $a > 0$). It is strictly positive for all real inputs. Whether $x$ is positive, negative, or zero, $e^{-x}$ will always be a positive value. It approaches zero as $x$ goes to infinity, but it never actually touches the x-axis. Because it is strictly positive, it cannot change the sign of our derivative; it only scales it. That is why we can safely ignore it when determining where the function increases or decreases!

Q: Why can't we just differentiate the terms separately? Like, derivative of $(x+2)$ times derivative of $e^{-x}$?

I know it is tempting to want to treat them independently, but calculus demands respect for the structure of the function. Because $(x+2)$ and $e^{-x}$ are multiplied together, they are 'entangled.' The Product Rule exists precisely because the rate of change of a product is not just the product of the rates of change. If you differentiate them separately, you miss the interaction terms that arise from the Product Rule. Always use the Product Rule for products!

Q: How do I know if I should use a test point or just look at the linear factor?

You can do either! Using a test point is a great way to verify your work, but analyzing the linear factor is faster and more elegant. Since we established that $e^{-x}$ is always positive, the sign of $f'(x)$ is identical to the sign of $-(x+1)$. If $x -1$, then $(x+1)$ is positive, so $-(x+1)$ is negative. It is a beautiful, logical flow that avoids the need for test points entirely once you are confident in your algebra.

Question

The function defined by $f(x) = (x + 2) e^{-x}$ is

Options:

(A) decreasing for all $x$

(B) decreasing in $(-\infty, -1)$ and increasing in $(-1, \infty)$

(C) increasing for all $x$

(D) decreasing in $(-1, \infty)$ and increasing in $(-\infty, -1)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers! Today, we are not just solving a problem; we are dissecting the behavior of a function. We are looking at $f(x) = (x + 2) e^{-x}$.

When you look at this function, I want you to see a story. You have a linear term, $(x+2)$, which wants to grow to infinity, and an exponential decay term, $e^{-x}$, which wants to crush everything down to zero.

The result of this interaction is a curve that climbs, peaks, and then slides away. Our mission is to find exactly where that transition happens.

### The Master Key: The First Derivative

In the world of calculus, the first derivative, $f'(x)$, is our compass. It tells us the direction of the slope.

If $f'(x) > 0$, the function is climbing (increasing). If $f'(x) < 0$, the function is sliding (decreasing).

To find this, we must apply the Product Rule. We have $u = (x+2)$ and $v = e^{-x}$. The rule states that the derivative is $u'v + uv'$. Let us execute this with precision:

$$f'(x) = \frac{d}{dx}(x+2) \cdot e^{-x} + (x+2) \cdot \frac{d}{dx}(e^{-x})$$

Calculating the individual parts, the derivative of $(x+2)$ is simply $1$. The derivative of $e^{-x}$ (using the Chain Rule) is $-e^{-x}$. Putting it all together, we get:

$$f'(x) = 1 \cdot e^{-x} + (x+2) \cdot (-e^{-x})$$

### The Elegance of Simplification

Now, do not rush the algebra. This is where many students stumble. We have a common factor of $e^{-x}$. Let us pull it out:

$$f'(x) = e^{-x} [1 - (x + 2)]$$

Simplifying the terms inside the bracket, we get $1 - x - 2$, which simplifies to $-x - 1$. Thus, our derivative is:

$$f'(x) = -(x + 1) e^{-x}$$

This expression is the heart of our problem. Because $e^{-x}$ is always strictly positive for all real $x$, it does not affect the sign of the derivative. The sign of $f'(x)$ is entirely dictated by the term $-(x+1)$.

### The Critical Moment

We set $f'(x) = 0$ to find our critical point. Since $e^{-x} 
eq 0$, we must have $-(x+1) = 0$, which gives us $x = -1$.

This is the peak of our mountain. To the left of $x = -1$, let us test a value like $x = -2$.

The term $-(x+1)$ becomes $-(-2+1) = -(-1) = 1$, which is positive. Thus, the function is **increasing on $(-\infty, -1)$**.

To the right of $x = -1$, let us test $x = 0$. The term $-(x+1)$ becomes $-(0+1) = -1$, which is negative.

Thus, the function is **decreasing on $(-1, \infty)$**.

### Conclusion

We have successfully mapped the terrain. The function climbs until it hits $x = -1$ and then descends forever.

This rigorous analysis confirms that the function is increasing in $(-\infty, -1)$ and decreasing in $(-1, \infty)$. Keep this logical process in your toolkit, and no function will ever be able to hide its secrets from you!

Animated Solution for Mathematics - Differentiation: The function defined by $f (x) = (x + 2) e^{- x}$ is

Select Answer:

Visualized Solution

Visualizing the Function $f (x)$

The Monotonicity Tool

Applying the Product Rule

Calculating the Derivative

Simplifying $f^{'} (x)$

Finding the Critical Point

Testing the Left Interval $(- \infty, - 1)$

Testing the Right Interval $(- 1, \infty)$

Final Conclusion

The Sigma Insight: Monotonicity

Analyzing the Setup

The Master Key

The Elegance of Simplification

The Critical Moment

Conclusion

Similar Questions

The function $f (x) = x e^{x (1 - x)}, x \in R$ , is

If $f (x) = x e^{x (1 - x)}$ , then $f (x)$ is

The function $f (x) = \frac{l n ( π + x )}{l n ( e + x )}$ is

Let $f : (- 1, \infty) \to R$ be defined by $f (0) = 1$ and $f (x) = \frac{1}{x} lo g_{e} (1 + x)$ , $x \neq = 0$ . Then the function $f$ :

Let $f : R \to R$ be defined as $f (x) = {- \frac{4}{3} x^{3} + 2 x^{2} + 3 x 3 x e^{x}, x > 0, x \leq 0$ . Then $f$ is increasing function in the interval

Let $f (x) = \int e^{x} (x - 1) (x - 2) d x$ . Then $f$ decreases in the interval

If $f (x) = \int_{x^{2}}^{x^{2} + 1} e^{- t^{2}} d t$ , then $f (x)$ increases in

The function $f (x) = \frac{x}{x ^{2} - 6 x - 16}, x \in R - {- 2, 8}$

Let $f$ be a real valued function, defined on $R - {- 1, 1}$ and given by $f (x) = 3 lo g_{e} \frac{x - 1}{x + 1} - \frac{2}{x - 1}$ . Then in which of the following intervals, function $f (x)$ is increasing?

If $f : R \to R$ is a differentiable function such that $f^{'} (x) > 2 f (x)$ for all $x \in R$ , and $f (0) = 1$ , then

Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The function defined by f(x)=(x+2)e−x is

Select Answer:

Visualized Solution

Visualizing the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

The Monotonicity Tool

Applying the Product Rule

Calculating the Derivative

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(x)

Finding the Critical Point

Testing the Left Interval .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (−∞,−1)

Testing the Right Interval .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (−1,∞)