Problem 34 - Thomas Calculus

Question

Find the function with the given derivative whose graph passes through the point \(P .\) \(g^{\prime}(x)=\frac{1}{x^{2}}+2 x, \quad P(-1,1)\)

Step-by-Step Solution

Answer

The function is \(g(x) = -\frac{1}{x} + x^2 - 1\).

Step 1: Find the antiderivative of the derivative

The first step is to integrate the given derivative function \(g'(x) = \frac{1}{x^2} + 2x\) to find the antiderivative \(g(x)\). This is achieved by integrating each term separately:\[\int \left(\frac{1}{x^2} + 2x\right) \, dx = \int \frac{1}{x^2} \, dx + \int 2x \, dx\]The integrals are:\[\int \frac{1}{x^2} \, dx = \int x^{-2} \, dx = -x^{-1} = -\frac{1}{x}\]\[\int 2x \, dx = x^2\]Combining these results, we have:\[g(x) = -\frac{1}{x} + x^2 + C\]where \(C\) is the constant of integration.

Step 2: Use the initial condition to find the constant C

We know that the function \(g(x)\) must pass through the point \((-1, 1)\), meaning \(g(-1) = 1\). We can substitute \(x = -1\) and \(g(x) = 1\) into the equation to find \(C\):\[g(-1) = -\frac{1}{-1} + (-1)^2 + C = 1\]This simplifies to:\[1 + 1 + C = 1\]\[2 + C = 1\]Solving for \(C\), we get:\[C = 1 - 2 = -1\]

Step 3: Write the function with the found constant C

Now that we have determined \(C = -1\), we can write the full expression for \(g(x)\):\[g(x) = -\frac{1}{x} + x^2 - 1\]

Step 4: Verify the function passes through the point P

To ensure accurateness, substitute \(-1\) back into the function \(g(x)\) to check if it returns 1:\[g(-1) = -\frac{1}{-1} + (-1)^2 - 1\]This simplifies to:\[1 + 1 - 1 = 1\]Since the equation holds true, the function is verified to pass through the point \((-1, 1)\).

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus that involves finding the rate at which a function is changing at any given point. It is the process of computing the derivative of a function, which represents an instantaneous rate of change. This is incredibly important in understanding how functions behave across different points.

• For a function \(f(x)\), its derivative \(f'(x)\) indicates how rapidly and in what direction the function \(f(x)\) is changing at any particular value of \(x\).
• The derivative itself is the function of \(x\) where the slope of the tangent line at any \(x\) position represents the value of \(f'(x)\).

Differentiation helps in solving rates of change problems, such as determining velocity or growth speed in various contexts. When solving a problem that involves finding a function from its derivative, like the exercise we discussed, we are essentially moving "backwards" from differentiation — a process known as finding the antiderivative or integration.

Integral Calculus

Integral calculus, largely concerned with finding antiderivatives, is essential for solving problems related to the area under curves, accumulations, and totals. In our exercise, integral calculus is used to recover the original function from its derivative.

• The process of integration is essentially the reverse of differentiation. It involves finding a function \(g(x)\) when its derivative \(g'(x)\) is provided.
• For instance, to integrate \(g'(x) = \frac{1}{x^2} + 2x\), we calculate the antiderivative as \(g(x) = -\frac{1}{x} + x^2 + C\), where \(C\) is the constant of integration.

Integration provides the comprehensive description of how quantities accumulate over intervals and how individual rates sum to global effects. Understanding integration deeply enriches the ability to conceptualize problems where cumulative changes are involved.

Initial Conditions

Initial conditions are vital in determining the specific function out of the family of possible antiderivatives. Once we carry out indefinite integration, it results in a general solution with a constant \(C\), representing an infinite set of functions that differ by a constant shift.

• Initial conditions, such as the point \((-1, 1)\) through which the function must pass in our exercise, are used to find this constant. By substituting \(x = -1\) and \(g(x) = 1\) into the integrated function, we solve for \(C\).
• These conditions ensure that the solution aligns with specific information or real-world constraints, tailoring the general solution to match particular scenarios.

Setting initial conditions helps narrow down the correct antiderivative from many possibilities. It's crucial for achieving precise and relevant solutions in applied mathematics and sciences.