Key Concepts and Formulas

• Monotonicity: A function $f(x)$ is strictly increasing if $f'(x) > 0$ for all $x$ in its domain, and strictly decreasing if $f'(x) < 0$ for all $x$ in its domain.
• Number of Real Roots: A strictly monotonic function can have at most one real root.
• Intermediate Value Theorem: If a continuous function $f(x)$ takes values $f(a)$ and $f(b)$ with $f(a) < 0$ and $f(b) > 0$, then there exists a $c$ between $a$ and $b$ such that $f(c) = 0$.
• Odd Degree Polynomials: Odd degree polynomials have limits of opposite signs as $x$ approaches positive and negative infinity.

Step-by-Step Solution

Step 1: Define the function We are given the equation ${x^7} + 5{x^3} + 3x + 1 = 0$. We define the function $f(x)$ as: $f(x) = x^7 + 5x^3 + 3x + 1$ Our goal is to find the number of real solutions to $f(x) = 0$.

Step 2: Calculate the first derivative, $f'(x)$ To analyze the monotonicity of $f(x)$, we need to find its first derivative, $f'(x)$. We use the power rule of differentiation: $\frac{d}{dx}(x^n) = nx^{n-1}$. $f'(x) = \frac{d}{dx}(x^7 + 5x^3 + 3x + 1) = \frac{d}{dx}(x^7) + \frac{d}{dx}(5x^3) + \frac{d}{dx}(3x) + \frac{d}{dx}(1)$ $f'(x) = 7x^6 + 15x^2 + 3 + 0$ $f'(x) = 7x^6 + 15x^2 + 3$

Step 3: Analyze the sign of $f'(x)$ We want to determine if $f'(x)$ is always positive, always negative, or changes sign. We observe that $x^6 \ge 0$ and $x^2 \ge 0$ for all real numbers $x$. Therefore, $7x^6 \ge 0$ and $15x^2 \ge 0$. $f'(x) = \underbrace{7x^6}_{\ge 0} + \underbrace{15x^2}_{\ge 0} + \underbrace{3}_{> 0}$ Since each term is non-negative, and the constant term is strictly positive, the sum must be strictly positive. Thus, $f'(x) > 0 \quad \text{for all } x \in \mathbb{R}$

Step 4: Determine the monotonicity of $f(x)$ Since $f'(x) > 0$ for all $x$, the function $f(x)$ is strictly increasing on the entire real line.

Step 5: Relate monotonicity to the number of real roots A strictly increasing function can intersect the x-axis at most once. Therefore, the equation $f(x) = 0$ can have at most one real solution.

Step 6: Analyze the function behavior as $x \to \pm \infty$ Since $f(x)$ is a polynomial of odd degree (degree 7), its end behavior is such that: As $x \to \infty$, $f(x) \to \infty$. As $x \to -\infty$, $f(x) \to -\infty$. Since $f(x)$ is a polynomial, it is continuous everywhere. By the Intermediate Value Theorem, there must be at least one real root.

Step 7: Check f(0) $f(0) = 0^7 + 5(0)^3 + 3(0) + 1 = 1 > 0$. Since the function is strictly increasing, and $f(0)$ is already positive, the function can never cross the x-axis. Therefore, there are no real roots.

Common Mistakes & Tips

• Sign Errors: Be careful when calculating the derivative and analyzing its sign.
• Monotonicity vs. Roots: A strictly monotonic function can have at most one root. However, it may not have any roots.
• Intermediate Value Theorem: Remember that the Intermediate Value Theorem only guarantees the existence of at least one root within an interval, not exactly one.

Summary We analyzed the function $f(x) = x^7 + 5x^3 + 3x + 1$ by finding its derivative $f'(x) = 7x^6 + 15x^2 + 3$. We determined that $f'(x) > 0$ for all real $x$, meaning $f(x)$ is strictly increasing. Since $f(0) = 1 > 0$, and the function is strictly increasing, $f(x)$ is always positive and never intersects the x-axis. Therefore, there are no real roots.

Final Answer The final answer is \boxed{0}, which corresponds to option (A).