Problem 60 - 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. $$ g(x)=\frac{1}{3} e^{-x} $$

Step-by-Step Solution

Answer

The function \( g(x) \) is always decreasing and concave up with no critical values or inflection points.

Step 1: Understand the Function

The given function is \( g(x) = \frac{1}{3} e^{-x} \). This is an exponential function where the base \( e \) is being raised to the negative variable \( x \), causing it to decay as \( x \) increases.

Step 2: Determine the Critical Values

To find critical values, we need to find the derivative of \( g(x) \) and set it equal to zero. The derivative is: \( g'(x) = -\frac{1}{3} e^{-x} \). Since the exponential function \( e^{-x} \) is always positive, \( g'(x) = 0 \) has no solutions. Therefore, there are no critical values.

Step 3: Analyze Increasing and Decreasing Intervals

Since \( g'(x) = -\frac{1}{3} e^{-x} \) is always negative, the function \( g(x) \) is decreasing for all \( x \in (-\infty, \infty) \).

Step 4: Determine the Concavity and Inflection Points

To find the concavity, calculate the second derivative: \( g''(x) = \frac{1}{3} e^{-x} \). Since this is always positive, the function is concave up for all \( x \in (-\infty, \infty) \). There are no inflection points since the concavity does not change.

Step 5: Graph the Function

Plot the graph of \( g(x) = \frac{1}{3} e^{-x} \). The graph starts at \( y = \frac{1}{3} \) when \( x = 0 \) and decreases asymptotically to zero as \( x \) approaches infinity. The curve is always concave up, reflecting the results from the second derivative.

Key Concepts

Exponential Function

Exponential functions are mathematical expressions where a constant base is raised to the power of a variable exponent. In the function \( g(x) = \frac{1}{3} e^{-x} \), the base \( e \) is Euler's number, approximately equal to 2.718. An important characteristic of exponential functions is how they grow or decay. When the exponent is negative, as in our function, it results in exponential decay, meaning the value of the function decreases as \( x \) increases.

This specific form, \( \frac{1}{3} e^{-x} \), begins at a value of \( \frac{1}{3} \) when \( x = 0 \), and approaches zero as \( x \) becomes very large. Exponential functions are commonly used in various fields such as biology, finance, and physics because they can model real-life phenomena like populations and interest growth or decay.

Critical Values

Critical values of a function are points where the derivative of the function equals zero or is undefined. They are important because they help identify potential maxima, minima, or saddle points. For the function \( g(x) = \frac{1}{3} e^{-x} \), we compute the derivative \( g'(x) = -\frac{1}{3} e^{-x} \).

The exponential part \( e^{-x} \) is always positive, which means the derivative \( g'(x) \) itself will never be zero because it always multiplies by a non-zero constant. Hence, there are no critical values for this function, and thus no turning points where the behavior of the function might shift from increasing to decreasing or vice versa.

Concavity

Concavity in functions describes the direction in which a curve bends. To determine concavity, we examine the second derivative. For our function \( g(x) = \frac{1}{3} e^{-x} \), the second derivative is \( g''(x) = \frac{1}{3} e^{-x} \). This is a key insight because the second derivative, like the first, always involves \( e^{-x} \), which remains positive for all real \( x \).

Because \( g''(x) > 0 \) across its entire domain, the function is always concave up. There are no inflection points in this case, as inflection points require a change in concavity, which does not occur here. Functions that are concave up generally look like a bowl facing upwards and in the case of exponential decay, create a smooth curve sloping downwards.

Graphing Functions

Graphing functions helps visualize their behavior over a range of values. For the function \( g(x) = \frac{1}{3} e^{-x} \), graphing offers a clear picture of its decreasing, asymptotic nature. At \( x = 0 \), the graph starts at a height of \( \frac{1}{3} \) and gradually approaches the x-axis as \( x \) moves towards infinity, never quite touching it.

Here are a few steps for graphing this function effectively:

• Identify key points, such as the initial value at \( x = 0 \), where \( g(x) = \frac{1}{3} \).
• Recognize that there is no horizontal shift; the curve simply moves downwards.
• Note that the asymptote is the x-axis itself, since the function gets closer to zero as \( x \) increases.
• Since the curve never changes concavity, the graph remains uniformly concave up.

Graphing this function is a straightforward matter of illustrating its decreasing trend, starting from the initial point and continuing to approach the asymptote in a smooth, downward curve.