Problem 56 - Calculus and its applications

Question

Graph each function. Then determine any critical values, inflection points, intervals over which the function is increasing or decreasing, and the concavity. $$ f(x)=e^{2 x} $$

Step-by-Step Solution

Answer

The function increases and is concave up everywhere; no critical points or inflection points exist.

Step 1: Graph the Function

The given function is exponential: \( f(x) = e^{2x} \). An exponential function grows rapidly as \( x \) increases. Its graph slopes upwards. To sketch it, plot several points: \( f(0) = 1 \), \( f(1) = e^2 \), \( f(-1) = e^{-2} \). Connect these points with a smooth curve that increases as \( x \) moves rightwards.

Step 2: Find the Derivative

To determine critical values, find the first derivative of \( f(x) = e^{2x} \). Use the chain rule: the derivative is \( f'(x) = 2e^{2x} \). This derivative shows the slope of the function.

Step 3: Determine Critical Values

Critical values occur where \( f'(x) = 0 \) or is undefined. Since \( 2e^{2x} > 0 \) for all \( x \), it never equals zero, and \( e^{2x} \) is defined everywhere. Thus, there are no critical points.

Step 4: Identify Intervals of Increase or Decrease

Since \( f'(x) = 2e^{2x} > 0 \) for all \( x \), the function is increasing on the interval \((-\infty, \infty)\). It never decreases since the derivative is always positive.

Step 5: Find the Second Derivative

The second derivative will help find concavity and inflection points. Differentiate \( f'(x) = 2e^{2x} \) to get \( f''(x) = 4e^{2x} \).

Step 6: Determine Concavity and Inflection Points

Since \( f''(x) = 4e^{2x} > 0 \) for all \( x \), the function is concave up everywhere on \((-\infty, \infty)\). There are no inflection points since the concavity does not change.

Key Concepts

Critical Points

Critical points of a function are crucial in understanding where a function reaches its local maxima or minima. These points are where the derivative of the function equals zero or where the derivative does not exist. For the function \( f(x) = e^{2x} \), we first find its derivative, \( f'(x) = 2e^{2x} \), which utilizes the chain rule of differentiation.

• The derivative, \( 2e^{2x} \), is always positive, meaning it never touches zero.
• Since \( 2e^{2x} > 0 \) for all values of \( x \), and the exponential function \( e^{2x} \) is defined for all \( x \), \( f'(x) \) is never nonexistent.

Thus, there are no points where the slope of the function equals zero or is undefined. Therefore, the function \( f(x) = e^{2x} \) has no critical points. This constant positivity in its derivative also indicates continuous growth, ruling out local peaks or valleys.

Concavity

Concavity is about the direction the function opens—either upwards or downwards—and is determined by the second derivative of the function. For an exponential function like \( f(x) = e^{2x} \), understanding concavity involves calculating the second derivative. We start again from the first derivative, \( f'(x) = 2e^{2x} \).

• The second derivative is \( f''(x) = 4e^{2x} \), which simplifies to four times the value of \( f(x) \).
• Since \( 4e^{2x} \) is always greater than zero, this indicates the graph is concave up for all \( x \).

A graph that is concave up looks like a "U" shape, although for \( f(x) = e^{2x} \), it's an ever-steepening curve up to infinity.
Since the second derivative doesn't change sign—it's always positive—there are no inflection points where the graph changes its concavity. This continuity ensures that the graph of \( f(x) \) is perpetually opening upwards.

Increasing and Decreasing Intervals

Analyzing where a function is increasing or decreasing helps us understand the behavior of the function's growth over the domain of \( x \). For \( f(x) = e^{2x} \), we use the first derivative, \( f'(x) = 2e^{2x} \), to determine intervals of increase or decrease.

• The derivative \( f'(x) = 2e^{2x} \) is always positive for all values of \( x \).
• This steady positivity means the function is continuously increasing over its entire domain \((-\infty, \infty)\).
• Since the derivative is never zero and never negative, the function has no intervals of decrease.

Thus, \( f(x) = e^{2x} \) rises endlessly and steadily, reflecting a hallmark property of exponential growth: fast escalation without retreat. Recognizing these intervals allows for predicting the function's overarching trend, showcasing a constant upward trajectory.