Function Derivative Analysis for f(1)

We are given the derivative of a function $f'(x) = \frac{8}{x^2 + 3x + 4}$ for $x > 0$, and the initial condition $f(0) = 1$. We need to find the interval in which $f(1)$ lies.

Bounding the Derivative f'(x)

Consider the denominator $g(x) = x^2 + 3x + 4$. To find its behavior on the interval $[0, 1]$, we check its values at the endpoints:

• At $x = 0$, $g(0) = 0^2 + 3(0) + 4 = 4$.
• At $x = 1$, $g(1) = 1^2 + 3(1) + 4 = 8$.

The derivative of the denominator is $g'(x) = 2x + 3$. For $x \in [0, 1]$, $g'(x)$ is positive, meaning $g(x)$ is increasing on this interval. Therefore, for $x \in [0, 1]$, the denominator satisfies:

$ 4 \le x^2 + 3x + 4 \le 8 $

Taking the reciprocal reverses the inequality:

$ \frac{1}{8} \le \frac{1}{x^2 + 3x + 4} \le \frac{1}{4} $

Multiplying by 8 gives the bounds for $f'(x)$ on $[0, 1]$:

$ \frac{8}{8} \le \frac{8}{x^2 + 3x + 4} \le \frac{8}{4} $ $ 1 \le f'(x) \le 2 $

Applying the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus states that $f(1) - f(0) = \int_{0}^{1} f'(x) dx$. Using the bounds we found for $f'(x)$:

$ \int_{0}^{1} 1 \, dx \le \int_{0}^{1} f'(x) \, dx \le \int_{0}^{1} 2 \, dx $

Evaluating the integrals:

$ [x]_{0}^{1} \le f(1) - f(0) \le [2x]_{0}^{1} $ $ (1 - 0) \le f(1) - f(0) \le (2(1) - 2(0)) $ $ 1 \le f(1) - f(0) \le 2 $

Determining the Interval for f(1)

We are given $f(0) = 1$. Substituting this value into the inequality:

$ 1 \le f(1) - 1 \le 2 $

Add 1 to all parts of the inequality to solve for $f(1)$:

$ 1 + 1 \le f(1) \le 2 + 1 $ $ 2 \le f(1) \le 3 $

Thus, $f(1)$ lies in the interval $[2, 3]$.