Problem 41 - Calculus: Early Transcendental Functions

Question

Find the function \(f(x)\) satisfying the given conditions. $$f^{\prime \prime}(x)=12, f^{\prime}(0)=2, f(0)=3$$

Step-by-Step Solution

Answer

The function that satisfies the given conditions is \(f(x) = 6x^{2} + 2x + 3\).

Step 1: Determine the First Derivative

To determine the first derivative \(f^{\prime}(x)\), you integrate the given second derivative \(f^{\prime \prime}(x)=12\) with respect to \(x\).\[\int f^{\prime \prime}(x) \, dx = \int 12 \, dx\]Doing the integration leads to:\[f^{\prime}(x) = 12x + C\]where \(C\) is the integration constant.

Step 2: Apply the Initial Condition for \(f^{\prime}(0)\)

Substitute \(x=0\) into \(f^{\prime}(x) = 12x + C\) and set it equal to the given initial condition \(f^{\prime}(0)=2\):\[2 = 12 \cdot 0 + C \]Solving for \(C\) produces \(C = 2\). So, the first derivative is \(f^{\prime}(x) = 12x + 2\).

Step 3: Determine the Original Function

To obtain the original function \(f(x)\), you integrate the first derivative \(f^{\prime}(x) = 12x + 2\) with respect to \(x\):\[\int f^{\prime}(x) \, dx = \int (12x + 2) \, dx\]Doing the integration gives:\[f(x) = 6x^{2} + 2x + K\]where \(K\) is another integration constant.

Step 4: Apply the Initial Condition for \(f(0)\)

Substitute \(x=0\) into \(f(x) = 6x^{2} + 2x + K\) and set it equal to the given initial condition \(f(0)=3\):\[3 = 6 \cdot 0^{2} + 2 \cdot 0 + K\]Solving for \(K\) produces \(K = 3\). So, the function that satisfies all the given conditions is \(f(x) = 6x^{2} + 2x + 3\).

Key Concepts

Integration

Integration is a fundamental tool in calculus that helps us find functions from their derivatives. It's the reverse process of differentiation. In this exercise, we used integration to determine the original function from its second derivative.
Here is how it works:

• You are given the second derivative \(f''(x) = 12\)
• By integrating \(f''(x)\), you find the first derivative \(f'(x)\).
• The integration of \(12\) with respect to \(x\) results in \(f'(x) = 12x + C\), where \(C\) is a constant.

This constant arises because integration can only determine a function up to an additive constant. To remove this ambiguity, initial conditions are used to solve for such constants.

Initial Conditions

Initial conditions are crucial in solving differential equations because they allow us to find specific solutions rather than a family of solutions.
Consider the integration process:

• We found \(f'(x) = 12x + C\).
• An initial condition \(f'(0) = 2\) helps us solve for \(C\).

By substituting \(x = 0\) into \(f'(x) = 12x + C\) and equating it to 2, we determined \(C = 2\). Similarly, another initial condition \(f(0) = 3\) was used to solve for another constant \(K\) in the original function \(f(x) = 6x^2 + 2x + K\).
This specific solution was uncovered by our initial conditions, providing us with \(f(x) = 6x^2 + 2x + 3\).

Calculus

Calculus is the branch of mathematics that studies change. It is divided mainly into two parts: differential calculus and integral calculus.
In this exercise, we used both aspects:

• Differential Calculus: We began with the problem stating the second derivative, \(f''(x) = 12\).
• Integral Calculus: Used integration to go step-by-step backwards, from the second derivative to the original function.

Differential calculus focuses on finding rates at which quantities change. In contrast, integral calculus allows us to compute quantities like areas under curves, which sum up these rates to reveal total amounts.
This harmony between differentiation and integration makes calculus a potent tool in mathematical problem-solving.

Second Derivative

The second derivative provides information about the curvature or concavity of a function.
For instance, a constant second derivative like \(f''(x) = 12\) tells us this function is parabolic:

• If the second derivative is positive, the function is concave up, resembling a smile.
• If it is negative, the function is concave down, resembling a frown.

In our example, due to the constant value of 12, the original function \(f(x) = 6x^2 + 2x + 3\) forms an upward-opening parabola.
The second derivative is a powerful tool that can provide insight into the behavior and characteristics of functions, making it a key topic in the study of calculus.