Key Concepts and Formulas

• Intermediate Value Theorem (IVT): If a continuous function $f(x)$ on an interval $[a, b]$ takes values $f(a)$ and $f(b)$ at the endpoints, then it also takes any value between $f(a)$ and $f(b)$ at some point within the interval.
• Derivatives and Monotonicity: If $f'(x) > 0$ on an interval, then $f(x)$ is increasing on that interval. If $f'(x) < 0$ on an interval, then $f(x)$ is decreasing on that interval. If $f''(x) > 0$ on an interval, then $f'(x)$ is increasing on that interval.
• Exponential and Logarithmic Functions: $e^x > 0$ for all real $x$. The function $x = \ln t$ is the inverse of $t = e^x$, and is defined for $t > 0$.

Step-by-Step Solution

1. Transformation to a Polynomial Equation

We are given the equation $e^{4x} + 2e^{3x} - e^x - 6 = 0$. To simplify this, we make the substitution $t = e^x$. Since $e^x$ is always positive for any real $x$, we have $t > 0$. Substituting $t$ into the equation, we get $(e^x)^4 + 2(e^x)^3 - e^x - 6 = 0$ $t^4 + 2t^3 - t - 6 = 0$ Let $f(t) = t^4 + 2t^3 - t - 6$. Our problem is now equivalent to finding the number of positive real roots of the polynomial equation $f(t) = 0$. Each positive value of $t$ corresponds to a unique real value of $x$ given by $x = \ln t$.

2. Analyzing the Monotonicity of $f(t)$ for $t > 0$ using the First Derivative

To understand the behavior of $f(t)$ and determine the number of positive roots, we first find the first derivative of $f(t)$ with respect to $t$: $f'(t) = \frac{d}{dt}(t^4 + 2t^3 - t - 6) = 4t^3 + 6t^2 - 1$ The first derivative $f'(t)$ tells us about the monotonicity of $f(t)$.

3. Analyzing the Monotonicity of $f'(t)$ for $t > 0$ using the Second Derivative

Next, we calculate the second derivative of $f(t)$ with respect to $t$: $f''(t) = \frac{d}{dt}(4t^3 + 6t^2 - 1) = 12t^2 + 12t$ The second derivative $f''(t)$ tells us about the concavity of $f(t)$, and more importantly, the monotonicity of $f'(t)$. For $t > 0$, we have $f''(t) = 12t^2 + 12t = 12t(t+1)$. Since $t > 0$, both $12t$ and $(t+1)$ are positive. Therefore, $f''(t) > 0$ for all $t > 0$. This means that $f'(t)$ is strictly increasing for all $t > 0$.

4. Finding Critical Points of $f(t)$

Since $f'(t)$ is strictly increasing for $t > 0$, it can have at most one root. Let's evaluate $f'(t)$ at some points:

• As $t \to 0^+$, $f'(t) \to -1$, so $f'(0) = -1$.
• $f'(1) = 4(1)^3 + 6(1)^2 - 1 = 4 + 6 - 1 = 9$. Since $f'(0) = -1 < 0$ and $f'(1) = 9 > 0$, and $f'(t)$ is continuous, by the Intermediate Value Theorem (IVT), there exists a root $t_0 \in (0, 1)$ such that $f'(t_0) = 0$. Furthermore, because $f'(t)$ is strictly increasing for $t > 0$, this root $t_0$ must be unique. Thus, $f(t)$ has a unique local minimum at $t = t_0$ for $t > 0$.

5. Analyzing the Behavior of $f(t)$ to Find its Roots

We know $f(t)$ decreases until $t_0$ and then increases. Let's evaluate $f(t)$ at some points:

• As $t \to 0^+$, $f(t) \to -6$.
• $f(1) = (1)^4 + 2(1)^3 - 1 - 6 = 1 + 2 - 1 - 6 = -4$.
• $f(2) = (2)^4 + 2(2)^3 - 2 - 6 = 16 + 16 - 2 - 6 = 24$. Since $f(1) = -4$ and $f(2) = 24$, and $f(t)$ is continuous, by the Intermediate Value Theorem, there must be at least one root in the interval $(1, 2)$. Since $t_0 \in (0,1)$, the function $f(t)$ is decreasing for $0 < t < t_0$ and then increasing for $t > t_0$. The minimum value $f(t_0)$ must be less than $f(1) = -4$. Therefore, $f(t_0) < -4 < 0$. Since the minimum value of $f(t)$ for $t > 0$ is negative ($f(t_0) < 0$), there must be exactly one positive root for $f(t)$. This root lies in the interval $(1, 2)$.

6. Relating Back to the Original Equation

We found that the polynomial $f(t) = 0$ has exactly one positive real root. Let this root be $t_R$. Since $t = e^x$, for each positive value of $t$, there is a unique real value of $x$ given by $x = \ln t$. Therefore, since there is exactly one positive root $t_R$ for $f(t)=0$, there is exactly one real root for the original equation $e^{4x} + 2e^{3x} - e^x - 6 = 0$.

Common Mistakes & Tips

• Domain of t: Remember that $t = e^x > 0$.
• IVT Application: Use the Intermediate Value Theorem to prove the existence of roots when a continuous function changes sign.
• Interpreting Derivatives: Understand the relationship between the sign of the first and second derivatives and the monotonicity and concavity of the function.

Summary

By substituting $t=e^x$, the exponential equation was transformed into a polynomial equation $f(t) = t^4 + 2t^3 - t - 6 = 0$. Analyzing the derivatives showed that $f''(t) > 0$ for $t > 0$, implying $f'(t)$ is strictly increasing. This led to the conclusion that $f'(t)$ has a unique positive root $t_0 \in (0,1)$. Consequently, $f(t)$ has a unique minimum at $t_0$. Evaluating $f(1) = -4$ and $f(2) = 24$ confirmed that this minimum is negative and that $f(t)$ crosses the t-axis exactly once for $t>0$. Thus, there is only one positive root for $f(t)$, which translates to exactly one real root for the original equation.

The final answer is $\boxed{1}$, which corresponds to option (C).