Key Concepts and Formulas

• Derivatives and Monotonicity: The sign of the first derivative, $f'(x)$, indicates whether a function $f(x)$ is increasing ($f'(x) > 0$) or decreasing ($f'(x) < 0$).
• Critical Points: Critical points occur where $f'(x) = 0$ or $f'(x)$ is undefined. These points are potential locations of local maxima or minima.
• Intermediate Value Theorem: If a continuous function $f(x)$ takes on values $f(a)$ and $f(b)$ at points $a$ and $b$, then it also takes on every value between $f(a)$ and $f(b)$ at some point between $a$ and $b$.

Step-by-Step Solution

Step 1: Define the Function and Objective

Our objective is to find the number of distinct real roots for the equation $x^4 - 4x + 1 = 0$. Let $f(x) = x^4 - 4x + 1$. We need to find how many values of $x$ make $f(x) = 0$. Why this step? Defining the function allows us to analyze it using calculus techniques.

Step 2: Calculate the First Derivative

To understand the function's behavior (where it increases or decreases), we compute its first derivative, $f'(x)$. $f'(x) = \frac{d}{dx}(x^4 - 4x + 1)$ $f'(x) = 4x^3 - 4$ Why this step? The sign of $f'(x)$ tells us whether $f(x)$ is increasing (if $f'(x) > 0$) or decreasing (if $f'(x) < 0$). Critical points, where $f'(x) = 0$ or is undefined, are potential locations for local maximum or minimum values, which help us map the function's shape.

Step 3: Find Critical Points

We set the first derivative equal to zero to find the critical points: $f'(x) = 0$ $4x^3 - 4 = 0$ $4(x^3 - 1) = 0$ $x^3 - 1 = 0$ $x^3 = 1$ $x = 1$

The only real solution is $x = 1$. Why this step? Identifying all real critical points is crucial for dividing the number line into intervals to analyze the function's monotonicity and locate potential turning points.

Therefore, the only real critical point for $f(x)$ is $x = 1$.

Step 4: Analyze Intervals of Increase and Decrease

We use the critical point $x=1$ to divide the real number line into two intervals: $(-\infty, 1)$ and $(1, \infty)$. We test the sign of $f'(x)$ in each interval to determine if $f(x)$ is increasing or decreasing.

• Interval $(-\infty, 1)$: Let's pick a test value, say $x=0$. $f'(0) = 4(0)^3 - 4 = -4$ Since $f'(0) < 0$, the function $f(x)$ is decreasing in the interval $(-\infty, 1)$.

• Interval $(1, \infty)$: Let's pick a test value, say $x=2$. $f'(2) = 4(2)^3 - 4 = 4(8) - 4 = 32 - 4 = 28$ Since $f'(2) > 0$, the function $f(x)$ is increasing in the interval $(1, \infty)$.

Why this step? This analysis tells us where the function is going "uphill" and "downhill", which helps us sketch its graph and find potential local extrema.

Step 5: Identify the Nature of Critical Points

At $x=1$, the function changes from decreasing to increasing. This indicates that $x=1$ is a point of local minimum. Why this step? Local minima and maxima are key features of a function's graph. If a local minimum is below the x-axis, it guarantees at least one real root on either side of that minimum.

Step 6: Evaluate Function at Extrema and Analyze End Behavior

Now, let's find the value of the function at this local minimum: $f(1) = (1)^4 - 4(1) + 1 = 1 - 4 + 1 = -2$ So, the function has a local minimum value of $-2$ at $x=1$. This means the point $(1, -2)$ is the lowest point in its immediate vicinity.

Next, we consider the end behavior of $f(x)$ as $x$ approaches positive and negative infinity:

• As $x \to \infty$, $f(x) = x^4 - 4x + 1$. The $x^4$ term dominates, so $f(x) \to \infty$.
• As $x \to -\infty$, $f(x) = x^4 - 4x + 1$. The $x^4$ term dominates, so $f(x) \to \infty$.

Why this step? Understanding the function's behavior at its extremities and its lowest/highest points helps us visualize its overall shape and predict how many times it will cross the x-axis.

Step 7: Determine the Number of Real Roots

We can now sketch the behavior of $f(x)$ based on our analysis:

1. As $x$ comes from $-\infty$, $f(x)$ starts from a very large positive value ($\infty$).
2. $f(x)$ decreases until it reaches its minimum value of $-2$ at $x=1$.
3. From $x=1$ onwards, $f(x)$ increases and goes towards $\infty$ as $x \to \infty$.

Since the minimum value of $f(x)$ is $-2$ (which is below the x-axis, i.e., $f(1) < 0$), and the function's values are positive and increasing towards infinity on both sides of the minimum, the graph of $f(x)$ must cross the x-axis exactly twice:

• Once in the interval $(-\infty, 1)$, where $f(x)$ decreases from $\infty$ to $-2$. By the Intermediate Value Theorem, there must be a root where $f(x)=0$.
• Once in the interval $(1, \infty)$, where $f(x)$ increases from $-2$ to $\infty$. By the Intermediate Value Theorem, there must be a root where $f(x)=0$.

These two crossings correspond to two distinct real roots.

Common Mistakes & Tips

• Tip: Always analyze the end behavior of the function to understand its overall shape. This helps confirm the number of real roots.
• Tip: Use the Intermediate Value Theorem to rigorously justify the existence of roots within intervals where the function changes sign.

Summary

The function $f(x) = x^4 - 4x + 1$ has a single local minimum at $(1, -2)$. Because this minimum is below the x-axis and the function tends towards positive infinity as $x \to \pm \infty$, the graph of $f(x)$ must intersect the x-axis at two distinct points. Therefore, there are two distinct real roots.

The final answer is \boxed{2}, which corresponds to option (B).