Problem Analysis: Finding f(2) for a Continuous Function

We are given a continuous function $f: (0,\infty) \to \mathbb{R}$ defined by:

• $f(x) = 2 + \frac{g(x)}{x}$ for $x > 0$
• $g(x) = \int_{1}^{x} f(t) dt$ for $x > 0$

Our goal is to find the specific value of $f(2)$.

Relating f(x) and g(x) using Calculus

From the definition of $g(x)$, we can use the Fundamental Theorem of Calculus Part 1 to find $g'(x)$:

$ g'(x) = \frac{d}{dx} \int_{1}^{x} f(t) dt = f(x) $

Now, substitute $g(x)$ and $g'(x)$ into the equation for $f(x)$. First, rewrite the relation $f(x) = 2 + \frac{g(x)}{x}$ as:

$ x f(x) = 2x + g(x) $

Differentiate both sides with respect to $x$, using the product rule for the left side:

$ \frac{d}{dx}(x f(x)) = \frac{d}{dx}(2x + g(x)) $

$ 1 \cdot f(x) + x \cdot f'(x) = 2 + g'(x) $

Substitute $g'(x) = f(x)$ into the equation:

$ f(x) + x f'(x) = 2 + f(x) $

Solving the Differential Equation for f(x)

Simplify the equation:

$ x f'(x) = 2 $

This is a simple differential equation. Separate the variables:

$ f'(x) = \frac{2}{x} $

Integrate both sides to find $f(x)$:

$ f(x) = \int \frac{2}{x} dx = 2 \int \frac{1}{x} dx = 2 \log_e |x| + C $

Since the domain is $(0, \infty)$, $x > 0$, so $|x| = x$. Thus:

$ f(x) = 2 \log_e x + C $

Determining the Constant of Integration C

We need to find the value of the constant $C$. We use the properties of $g(x)$:

$ g(1) = \int_{1}^{1} f(t) dt = 0 $

Substitute $x=1$ into the original relation $f(x) = 2 + \frac{g(x)}{x}$:

$ f(1) = 2 + \frac{g(1)}{1} = 2 + \frac{0}{1} = 2 $

Now, use this value in the expression for $f(x)$:

$ f(1) = 2 \log_e 1 + C $

$ 2 = 2 \cdot 0 + C $

$ C = 2 $

So, the function is $f(x) = 2 \log_e x + 2$.

Calculating the Final Value f(2)

Substitute $x=2$ into the function $f(x)$:

$ f(2) = 2 \log_e 2 + 2 $

We can rewrite $2 \log_e 2$ using logarithm properties ($\log_e a^b = b \log_e a$):

$ 2 \log_e 2 = \log_e (2^2) = \log_e 4 $

Therefore:

$ f(2) = 2 + \log_e 4 $