To find the fundamental period $T$ of $f(x)$, we analyze each term of the function:

$f(x) = \{2x^2 + 3x + \frac{5}{7}\} + \cos\left(6\pi x + \frac{\pi}{4}\right) + \sin\left(\frac{9\pi}{8} x\right) + \{x + \sqrt{2}\}.$

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1. Analysis of Each Term:

Term 1: $\{2x^2 + 3x + \frac{5}{7}\}$

The fractional part function, $\{y\} = y - \lfloor y \rfloor$, is periodic if $y$ is a periodic function. However:

• $2x^2 + 3x + \frac{5}{7}$ is a quadratic polynomial.
• A quadratic polynomial is not periodic because it grows indefinitely as $x \to \pm\infty$.
• Hence, $\{2x^2 + 3x + \frac{5}{7}\}$ does not contribute to periodicity.

Term 2: $\cos(6\pi x + \frac{\pi}{4})$

The cosine function is periodic, and its period depends on the coefficient of $x$: $\cos(6\pi x + \frac{\pi}{4}) \quad \text{has period} \quad T = \frac{2\pi}{6\pi} = \frac{1}{3}.$

Term 3: $\sin\left(\frac{9\pi}{8}x\right)$

The sine function is periodic, with its period determined by the coefficient of $x$: $\sin\left(\frac{9\pi}{8}x\right) \quad \text{has period} \quad T = \frac{2\pi}{\frac{9\pi}{8}} = \frac{16}{9}.$

Term 4: $\{x + \sqrt{2}\}$

• The fractional part of $x + \sqrt{2}$ is periodic if and only if $x + \sqrt{2}$ is periodic. However, since $\sqrt{2}$ is irrational, $\{x + \sqrt{2}\}$ is not periodic.

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2. Combine the Periodic Terms

The only periodic terms in $f(x)$ are: $\cos(6\pi x + \frac{\pi}{4}) \quad \text{with period } \frac{1}{3}, \quad \text{and} \quad \sin\left(\frac{9\pi}{8}x\right) \quad \text{with period } \frac{16}{9}.$ To find the fundamental period, $T$, we compute the least common multiple (LCM) of the periods $\frac{1}{3}$ and $\frac{16}{9}$.

Step 1: Express Periods as Fractions

$\frac{1}{3} = \frac{3}{9}, \quad \frac{16}{9}.$

Step 2: LCM of Numerators

The LCM of the numerators $3$ and $16$ is $48$. Thus, the LCM of $\frac{3}{9}$ and $\frac{16}{9}$ is: $T = \frac{\text{LCM}(3, 16)}{\text{GCD of denominators}} = \frac{48}{9}.$

Step 3: Simplify

[ \boxed! SOLLVE. ***