Question

Find an equation for the function \(f\) that has the indicated derivative and whose graph passes through the given point. $$ f^{\prime}(x)=0.4^{x / 3}, \quad\left(0, \frac{1}{2}\right) $$

Short answer

The equation for the required function is \(f(x) = \frac{1}{3\ln(0.4)} \cdot (0.4^{x / 3}-1) + 1/2\)

Step-by-step solution

Compute the Antiderivative

The first step is to find the antiderivative (also known as integral) of \(f'\). In this case, \(f'(x) = 0.4^{x / 3}\). This requires us to use the power rule for integration, which states that the integral of \(a^x\), where \(a\) is a constant, is \(\frac{a^x}{\ln(a)} + C\), where \(C\) is the constant of integration. So, the integral of \(0.4^{x / 3}\) is \(\frac{1}{3\ln(0.4)} \cdot 0.4^{x/3} + C\). We simplify this to be \(f(x) = \frac{1}{3\ln(0.4)} \cdot 0.4^{x / 3} + C\).

Use the given point to find the constant

The point \((0, 1/2)\) is on the function's graph, so we substitute \(x = 0\) and \(f(x) = 1/2\) into our equation to solve for \(C\). The equation becomes \(1/2 = \frac{1}{3\ln(0.4)} \cdot 0.4^{0} + C\), which further simplifies to \(1/2 = \frac{1}{3\ln(0.4)} + C\). Solving this equation gives \(C = 1/2 - \frac{1}{3\ln(0.4)}\).

Formulate the original function

Now we substitute \(C\) back into our function from Step 1. This gives \(f(x) = \frac{1}{3\ln(0.4)} \cdot 0.4^{x / 3} + 1/2 - \frac{1}{3\ln(0.4)}\), which simplifies to \(f(x) = \frac{1}{3\ln(0.4)} \cdot (0.4^{x / 3}-1) + 1/2\).

Key concepts

Integration by Power Rule

Grasping the integration by power rule is essential when you want to find the antiderivative of a function involving exponents. In the most basic form, the power rule for integration tells us that for a function of the form \( x^n \), where \( n \) is a number other than -1, the antiderivative is obtained by adding 1 to the exponent and then dividing by the new exponent, plus the constant of integration (\( C \)).

When we deal with exponential functions like \( a^{x} \), the rule adapts slightly. If \( a \) is a constant and \( a > 0 \), the power rule states that the integral of \( a^x \) is \( \frac{a^x}{\ln(a)} + C \). This is because the derivative of \( a^x \) brings down a \( \ln(a) \) from the exponent, and thus, we compensate for this in the integration process.

Using this rule, we can integrate functions like the given \( 0.4^{x / 3} \) with ease. The constant \( 0.4 \) is raised to an exponent function of \( x \), making this a perfect candidate for the power rule's extended application for exponential functions.

Constant of Integration

Whenever you compute the integral of a function, it's crucial to add a 'constant of integration', often denoted as \( C \). This is because when you derive a function, any constant term becomes zero, meaning that all antiderivatives actually represent a family of functions with differing constant terms.

Taking our example, after applying the integration by power rule to find the antiderivative of \( f'(x) = 0.4^{x / 3} \), we must add \( C \) to represent all possible functions that could have derived from the original function. This constant of integration is crucial for the next step: matching the antiderivative to a particular curve that corresponds to specific initial conditions or known points.

In practice, we find the value of \( C \) by using given points, such as \( (0, \frac{1}{2}) \) in this exercise. By substituting these point values into the antiderivative formula, we solve for \( C \), hence pinning down the specific function within the broader family of antiderivatives.

Exponential Functions

Exponential functions, such as the one in the exercise \( f'(x) = 0.4^{x / 3} \), are crucial in various fields due to their unique properties. Unlike polynomial functions which grow at rates determined by their degree, exponential functions grow at a rate proportional to their current value, which can result in rapid increases or decreases.

An exponential function is generally expressed as \( a^{x} \), where \( a \) is a positive constant, and \( x \) is the exponent. They are commonly used to model phenomena that change at a rate proportional to their current value, such as in population growth, radioactive decay, compound interest, and more.

Integral calculus involving exponential functions allows us to solve problems related to total growth or decay over time and can provide a function representing the quantity at any given time. The integral of the exponential function provided in the initial problem reflects this ability to account for continuous growth, with the power rule guiding the calculation process.