Question

If \(\mathrm{f}\) is a continuous function such that \(\mathrm{f}(\mathrm{xy})=\mathrm{f}(\mathrm{x}) . \mathrm{f}(\mathrm{y})\) for all positive real numbers \(\mathrm{x}\) and \(y\), then prove that \(f(x)=x^{k}\) for some constant \(\mathrm{k}\).

Short answer

Question: Prove that if a function f(x) is continuous and satisfies the condition f(xy) = f(x)f(y) for all positive real numbers x and y, then f(x) = x^k for some constant k. Answer: We've proved that f(x) = x^k for some constant k by following these steps: 1. Showed that f(1) = 1. 2. Found that f(x^2) = f(x)^2. 3. Demonstrated that f(x/y) = f(x)/f(y) for any positive x and y. 4. Established that f(x) > 0 for all x > 0. 5. Used logarithms to show that f(x) = x^k for a constant k.

Step-by-step solution

Find the value of f(1)

Let's first substitute x = 1 and y = 1 in the given equation f(xy) = f(x)f(y): f(1*1) = f(1)f(1) f(1) = f(1)^2 Since f is continuous, we know that f(1) is nonzero. Thus, we can divide by f(1) to get: f(1) = 1

Find the value of f(x^2)

Next, let's find the value of f(x^2) by substituting y = x: f(x*x) = f(x)f(x) f(x^2) = f(x)^2

Show f(x/y) = f(x)/f(y)

Now, let's find the value of f(x/y). Let z = x/y. Then, x = z * y. Substituting this in the given equation, we get: f(z*y) = f(z)f(y) Since we've already shown that f(xy)=f(x)f(y), we have: f(x) = f(z)f(y) Now, divide both sides by f(y) (since f(y) is nonzero for all positive y) to obtain: f(x)/f(y) = f(x/y) for any positive x and y

Show f(x) > 0 for all x > 0

We have f(x^2) = f(x)^2. Since the square of any real number is nonnegative, we obtain that f(x^2) ≥ 0. From the property f(xy)=f(x)f(y), we also know that f(x) is nonzero. Thus, f(x) > 0 for all x > 0.

Prove f(x) = x^k for a constant k

Now that we've shown f(x) > 0, we can use logarithms. Let g(x) = ln(f(x)). Then apply ln to both sides of the equation f(xy) = f(x)f(y): ln(f(xy)) = ln(f(x)f(y)) g(xy) = g(x) + g(y) for any x > 0 Let h(x) = g(e^x). Then we have: h(e^x * e^y) = h(e^x) + h(e^y) for any real x and y. Define k = h'(0), and note that k is a constant. Using the definition of the derivative, we have: k = lim[h(e^x) - h(1)] / x when x approaches 0 Since h is continuous, the limit exists. Now, we have h'(x) = lim[k * x] / x when x approaches 0. As x -> 0, k * x -> 0, so h'(x) = k for all x. Integrating both sides with respect to x, we obtain: h(x) = kx + c g(x) = k * ln(x) + c for the original function g Now let us find the value of c: g(1) = k * ln(1) + c 0 = c (since we have obtained f(1) = 1 and g(x) = ln(f(x))) So, g(x) = k * ln(x). Now, let's compute f(x) by exponentiating both sides: f(x) = e^(k * ln(x)) f(x) = x^k, where k is a constant Hence, we've proved that f(x) = x^k for some constant k.

Key concepts

Understanding Continuous Functions

Continuous functions are foundational in calculus, providing a smooth behavior of graphs and assuring no sudden jumps or breaks. To visualize continuity, imagine drawing the graph of a function without lifting your pen from the paper.

A formal definition involves limits: a function f is continuous at a point x if \(\lim_{{x \to c}} f(x) = f(c)\). When a function is continuous at every point within an interval, it is simply called continuous on that interval.

In the case of the function in the original exercise, continuity ensures that f(1) can be resolved with confidence to be 1, establishing a crucial step in the solution process. The ability to perform operations such as solving for f(1), finding f(x/y), and ultimately, using logarithms hinges on the function being continuous at all points.

Exponential Functions Explored

Exponential functions, yet another cornerstone of calculus, are identified by their distinct property: the rate of growth of these functions is proportional to the value of the function. A typical form of an exponential function can be written as \(f(x) = b^x\), where b is a positive real number called the base.

The number e, approximately equal to 2.71828, is a special base that often arises naturally in various scientific and financial calculations. In our textbook problem, the exponential function comes into play via the logarithms used to transform the original functional equation into one involving the natural logarithm, which has the base e. The beauty of the exponential function is its self-replicating nature under differentiation, which largely simplifies both theoretical and practical applications, including the proving of the result \(f(x) = x^k\).

Logarithmic Functions: Unlocking the Solution

Logarithmic functions are the inverse operations of exponential functions and crucial for solving various equations in calculus. The most common logarithm is the natural logarithm, denoted as \(\ln(x)\), which is the inverse of the exponential function with base e.

In our exercise, logarithms allow us to convert multiplication into addition, turning the complex function \(f(xy)\) into a linear form that can be differentiated and integrated. The function \(g(x) = \ln(f(x))\) translates the original functional equation into a form that assists in uncovering the underlying power relationship \(f(x) = x^k\). Understanding logarithms is vital: they enable us to move between levels of mathematical complexity, bridging components like multiplication and exponentiation to addition and multiplication, as exemplified in the systematic steps of the textbook solution.