Question

In each case, find a function \(f\) satisfying the given conditions. (a) \(\mathrm{f}\left(\mathrm{x}^{2}\right)=1 / \mathrm{x}\) for \(\mathrm{x}>0, \mathrm{f}(1)=1\). (b) \(f^{\prime}\left(\sin ^{2} x\right)=\cos ^{2} x\) for all \(x, f(1)=1\). (c) \(f^{\prime}(\sin x)=\cos ^{2} x\) for all \(x, f(1)=1\). (d) \(f^{\prime}(\ln x)=\left\\{\begin{array}{lll}1 & \text { for } & 01,\end{array}, f(0)=0\right.\).

Short answer

Q: Find a function f(x) that satisfies the given conditions in each part: a) f(x^2) = 1/x for x > 0, and f(1) = 1. b) f'(sin^2(x)) = cos^2(x) for all x, and f(1) = 1. c) f'(sin(x)) = cos^2(x) for all x, and f(1) = 1. d) f'(ln(x)) = 1 if 0 < x ≤ 1 and x if x > 1, while f(0) = 0. A: a) The function f(x) that satisfies the given conditions is f(x) = √(1/x). b) The function f(x) that satisfies the given conditions is f(x) = (1/4)sin(2sin^(-1)(√x)) + 1. c) The function f(x) that satisfies the given conditions is f(x) = sin(x) - (1/3)sin^3(x) + (1 - sin(1) + (1/3)sin^3(1)). d) The function f(x) that satisfies the given conditions is: f(x) = { x for 0 < x ≤ 1, (1/2)x^2 for x > 1. }

Step-by-step solution

Part (a)

First we need to find a function f(x) that satisfies f(x^2) = 1/x for x > 0. To do this, let f(x) be the square root of 1/x, meaning f(x) = \sqrt{1/x}, so that f(x^2) = \sqrt{1/x^2} = 1/x. Now we should check if f(1) = 1: f(1) = \sqrt{1/1} = 1, which is true. Thus, the function f(x) = \sqrt{1/x} satisfies the given conditions.

Part (b)

The given condition states that the derivative of f with respect to sin^2(x) is equal to cos^2(x). We should try to integrate cos^2(x) with respect to sin^2(x) to find f(x). To do this, we can first replace sin^2(x) with a variable u: u = sin^2(x), then du/dx = 2sin(x)cos(x). Now we can rewrite the condition as: df/du = cos^2(x) and integrate both sides with respect to u: 1. Find the integral of cos^2(x) with respect to sin^2(x): \(∫ \frac{df}{du} du = ∫ \cos^2(x) * \frac{du}{dx} dx\) \(∫ df = ∫ \frac{1}{2} \cos(2x) dx\) \(f(u) = \frac{1}{4} \sin(2x) + C\) 2. Replace sin^2(x) with u: \(f(sin^2(x)) = \frac{1}{4} \sin(2x) + C\) 3. We were also given f(1) = 1, so substituting into the function, we have \(1 = \frac{1}{4} \sin(2 * \sin^{-1}(\sqrt{1})) + C\) \(1 = \frac{1}{4} \sin(\pi) + C\) \(1 = C\) So the function \(f(x) = \frac{1}{4} \sin(2\sin^{-1}(\sqrt{x})) + 1\) satisfies the given conditions.

Part (c)

The given condition in this case is f'(sin(x)) = cos^2(x) for all x and f(1) = 1. Let's try integrating cos^2(x) with respect to sin(x) to find the function f(x): 1. Define a variable u = sin(x), which implies du/dx = cos(x): \(df/du = cos^2(x)\) \(df/du = (1-u^2)\) 2. Integrate both sides with respect to u: \(f(u) = ∫ (1-u^2) du\) \(f(u) = u - \frac{1}{3}u^3 + C\) \(f(sin(x)) = sin(x) - \frac{1}{3}sin^3(x) + C\) 3. Use f(1) = 1 to solve for C: \(f(sin(1)) = 1 = sin(1) - \frac{1}{3}sin^3(1) + C\) \(C = 1 - sin(1) + \frac{1}{3}sin^3(1)\) The function that satisfies the given conditions is \(f(x) = sin(x) - \frac{1}{3}sin^3(x) + (1 - sin(1) + \frac{1}{3}sin^3(1))\)

Part (d)

For this case, we are given that the derivative of f with respect to ln(x) is 1 if 0 < x ≤ 1, and x if x > 1. We can define f'(ln(x)) as g(x) where g(x) = 1 for 0 < x ≤ 1, and g(x) = x for x > 1. 1. Integrate g(x) with respect to x: \(f(\ln(x)) = ∫ g(x) dx\) \(f(\ln(x)) = ∫ (1) dx\) for 0 < x ≤ 1 \(f(\ln(x)) = ∫ (x) dx\) for x > 1 2. Solve the integrals: For 0 < x ≤ 1: \(f(\ln(x)) = x + C_1\) For x > 1: \(f(\ln(x)) = \frac{1}{2}x^2 + C_2\) 3. We also know f(0) = 0, so use this to solve for C_1 and C_2: For 0 < x ≤ 1: \(f(\ln(x)) = x\) For x > 1: \(f(\ln(x)) = \frac{1}{2}x^2\) The function that satisfies the given conditions is: $f(x) = \left\\{ \begin{array}{ll} x & \text{ for } 0 < x ≤ 1 \\ \frac{1}{2}x^2 & \text{ for } x > 1 \end{array} \right.$

Key concepts

Derivative of a Function

When studying Integral Calculus, understanding the concept of the derivative is essential. A derivative represents the rate at which a function changes at any given point. It is a fundamental tool for describing motion, growth, and many other types of changes. For example, consider part (c) of our exercise where the derivative of function f with respect to the sine of x is given as \( f'(\sin(x)) = \cos^2(x) \). This tells us how the function f is changing when we input the sine of x.

To understand derivatives better, you can think of them as the slope of a curve at a point. If the curve represents a hill, the derivative tells us how steep the hill is at that location. When solving problems like these in our exercise, being able to identify and work with derivatives allows us to find functions matching specific rates of change, which is exactly what we do when we find the antiderivative or the original function.

Integration Techniques

Integration is the reverse process of differentiation, and there are various integration techniques to solve complex integrals. In our exercise, we use substitution as a technique to simplify the integration process. For instance, in part (b), we substitute \( u = \sin^2(x) \) to find the antiderivative of the function \( f'(\sin^2(x)) = \cos^2(x)\).

Moreover, we often encounter functions that are not directly integrable in their original form, so we manipulate the equation to a more manageable form, as we did by using trigonometric identities. Sometimes, knowing which technique to apply and how to simplify the function can make the difference between an intractable problem and one that is easily solved. It's also crucial to remember that integration always includes an arbitrary constant \(C\), except when initial conditions are given that allow us to solve for it.

Initial Conditions

In many calculus problems, such as those we encounter in our exercise, we are given initial conditions. These conditions are used to determine the constant of integration when finding the general solution to a differential equation. Initial conditions essentially specify the value of the function or its derivatives at a particular point.

By applying these conditions to our general solution, we narrow down the infinite set of possible functions to the one specific function that passes through a given point. For instance, in part (a), using the initial condition \( f(1) = 1 \) allows us to determine the constant and thus find the unique function that satisfies both the differential equation and the initial condition. Initial conditions are vital for ensuring the resulting function has practical and precise application to the problem at hand.