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\(23-46\) Find \(f\) $$f^{\prime \prime}(x)=2+\cos x, \quad f(0)=-1, \quad f(\pi / 2)=0$$

Short Answer

Expert verified
The function is \(f(x) = x^2 - \cos x - (\pi/2)x\).

Step by step solution

01

Integrate to Find the First Derivative

To find the first derivative of the function, integrate the given second derivative with respect to \(x\): \[ f'(x) = \int (2 + \cos x) \, dx = 2x + \sin x + C_1 \]Where \(C_1\) is an integration constant that needs to be determined.
02

Find the Function by Integrating Again

Integrate the first derivative to find the function \(f(x)\):\[ f(x) = \int (2x + \sin x + C_1) \, dx = x^2 - \cos x + C_1x + C_2 \]Here, \(C_2\) is another integration constant.
03

Apply Initial Condition \(f(0) = -1\)

Use the initial condition \(f(0) = -1\) to find \(C_2\):\[ f(0) = 0^2 - \cos(0) + C_1(0) + C_2 = -1 \]\[ -1 + C_2 = -1 \]\[ C_2 = 0 \]
04

Apply Second Condition \(f(\pi/2) = 0\)

Substitute \(\pi/2\) into the function to find \(C_1\):\[ f(\pi/2) = (\pi/2)^2 - \cos(\pi/2) + C_1(\pi/2) = 0 \]\[ \pi^2/4 + C_1(\pi/2) = 0 \]Solving for \(C_1\):\[ C_1\pi/2 = -\pi^2/4 \]\[ C_1 = -\pi/2 \]
05

Write the Final Function

Now that we have \(C_1\) and \(C_2\), the function \(f(x)\) is:\[ f(x) = x^2 - \cos x - (\pi/2)x \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integration
Integration is a fundamental concept in calculus. It allows us to find a function given its derivative. During integration, we are essentially summing continuous values. In the context of an initial value problem, we're reversing differentiation. This particular maths problem started with a known second derivative
  • The second derivative, given as \( f''(x) = 2 + \cos x \), needed to be integrated to uncover the first derivative, \( f'(x) \).This yielded \( f'(x) = 2x + \sin x + C_1 \), with \( C_1 \) being an unknown constant.
  • Integrating once more yields the original function \( f(x) \) and introduces a second constant, \( C_2 \).
Integration requires knowledge of anti-derivatives. Ideally, a solid grasp on trigonometric identities and power rules is needed to carry out integration correctly. It's like reversing the math steps to build the whole picture again.
Initial Value Problem
An initial value problem is a differential equation paired with initial conditions. These conditions help determine constants of integration. Given the complexities of differential equations, initial conditions make the solution unique.
In our exercise, we started with two specific initial conditions. These were:
  • \( f(0) = -1 \)
  • \( f(\pi / 2) = 0 \)
To solve equations like these, the initial conditions are used post-integration to solve for the constants. They serve as key anchors to the problem, ensuring that once integrated, we arrive at the precise function that fits all initial conditions. By substituting these conditions, we systematically found the constants \( C_1 \) and \( C_2 \). This process gives us an exact function solution instead of a family of possible solutions.
Second Derivative
The second derivative represents the curvature or concavity of a function. It gives information on how the rate of change itself is changing. In simpler terms, if the first derivative describes velocity, the second derivative tells us about acceleration.
Our given mathematical problem started with the second derivative \( f''(x) = 2 + \cos x \). From here:
  • We integrated to find \( f'(x) \), which relates more to slopes or first-order change.
  • The problem then led us to \( f(x) \), the original function corresponding to a cumulative effect of those changes.
The second derivative often guides us in achieving the larger context of a function. It's a tool for understanding the shape and behavior of curves, such as identifying concave up/down intervals. This "peek behind the curtain" into the function's structure is extremely useful in math and science applications related to motion, growth, and more.

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