Problem 53 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Find the function \(f\) given that the slope of the tangent line to the graph of \(f\) at any point \((x, f(x))\) is \(f^{\prime}(x)\) and that the graph of \(f\) passes through the given point. $$f^{\prime}(x)=-2 x e^{-x^{2}+1} ;(1,0)$$

Step-by-Step Solution

Answer

The function \(f(x)\) is given by \(f(x) = e^{-x^2 + 1} - 1\).

Step 1: Integrate the derivative to find the original function

: We are given that the derivative of \(f(x)\) is \(f^{\prime}(x) = -2xe^{-x^2+1}\). To find the original function, we need to integrate the derivative with respect to \(x\). $$\int f^{\prime}(x) \,dx= \int -2xe^{-x^2+1} \,dx$$ To solve this integral, we can perform a substitution: let \(u = -x^2 + 1 \implies du = -2x \, dx\). So the integral becomes: $$\int -2xe^{-x^2+1} \, dx = \int e^u \, du$$ Now, we can integrate easily: $$\int e^u \, du = e^u + C$$ Now, substitute back \(u = -x^2 + 1\): $$f(x) = e^{-x^2 + 1} + C$$

Step 2: Find the constant of integration using the given point

: Now we need to find the value of the constant C using the given point \((1, 0)\) that the graph of \(f\) passes through. We know that if the point is on the graph of the function at \(x=1\), the function value should be \(f(1)=0\). We can use this to find the constant C: $$0 = e^{-(1)^2 + 1} + C$$ $$0 = e^{-1 + 1} + C$$ $$0 = e^0 + C$$ $$0 = 1 + C$$ $$C = -1$$

Step 3: Write the final solution

: Now that we have found the value of the constant C, we can write down the final solution. $$f(x) = e^{-x^2 + 1} - 1$$ So, the function \(f(x)\) is given by \(f(x) = e^{-x^2 + 1} - 1\).

Key Concepts

Differential Equations

Differential equations are fundamental in calculus and mathematics as a whole. They involve equations that relate a function with its derivatives. In simpler terms, differential equations show the relationship between a function and the rates at which it changes. They help us describe various phenomena such as heat conduction, fluid flow, and motion, using mathematical models.
For example, in the exercise, we're given a differential equation in the form of the derivative of the function, i.e., \(f'(x) = -2xe^{-x^2+1}\). This equation tells us that the slope of the tangent to the curve at any point \((x, f(x))\) is determined by this expression. Solving differential equations often involves finding the original function from its derivative.

Integration Techniques

Integration is a method used to find functions when their derivatives are known, essentially reversing differentiation. One of the key integration techniques used in calculus is the substitution method. Substitution is particularly handy when direct integration isn't feasible.
In the exercise, after identifying the derivative \(f'(x) = -2xe^{-x^2+1}\), we used a substitution to solve the integral. By letting \(u = -x^2 + 1\), we transformed the integral to a simpler form \(\int e^u \, du\). Then, we integrated to get \(e^u + C\), which simplifies back to \(e^{-x^2 + 1} + C\) upon substituting for \(u\). This process is fundamental for solving more complex integrals that arise from differential equations.

Tangent Line Slope

The slope of a tangent line gives us a snapshot of how a function behaves at a specific point. It's essentially the derivative of the function at that point. When you have the slope of the tangent line, you're equipped to understand the rate of change at any moment.
In our exercise, the function \(f'(x)\) represents this slope at every point on the curve \(f\). By integrating, we recovered the full description of the function, allowing us to understand not just the slope but the shape of the curve itself. Knowing the slope offers insights into increasing and decreasing behavior of \(f(x)\), crucial for understanding the graph of the function.

Constant of Integration

When integrating, a 'constant of integration' \(C\) arises because the process of differentiation erases constants. This constant accounts for all the possible vertical shifts of the function that would still have the same derivative. Thus, different values of \(C\) mean different functions that are parallel shifts of one another.
In the exercise, after calculating the integral, we had \(f(x) = e^{-x^2 + 1} + C\). To find the correct value of \(C\), we used the given point \((1, 0)\). Inputting \(x = 1\) into the equation yielded \(C = -1\), fixing the position of the function graph to pass through the point \((1, 0)\). This step is crucial whenever initial conditions or specific points are provided, ensuring the solution fits the specific scenario described in the problem.