Key Concepts and Formulas

• Finding Roots: The roots of a function $f(x)$ are the values of $x$ for which $f(x) = 0$. These correspond to the points where the graph of $y = f(x)$ intersects the x-axis.
• Derivatives and Critical Points: Critical points of a function $f(x)$ occur where $f'(x) = 0$ or $f'(x)$ is undefined. These points are potential locations of local maxima, local minima, or saddle points.
• Intermediate Value Theorem (IVT): If $f(x)$ is a continuous function on the interval $[a, b]$, and $k$ is any number between $f(a)$ and $f(b)$, then there exists at least one number $c$ in the interval $(a, b)$ such that $f(c) = k$. In particular, if $f(a)$ and $f(b)$ have opposite signs, then there exists at least one root in the interval $(a, b)$.

Step-by-Step Solution

Step 1: Find the first derivative of the function.

We are given the function $y = f(x) = x^5 - 20x^3 + 50x + 2$. To find the critical points, we need to find the first derivative $f'(x)$. $f'(x) = \frac{d}{dx}(x^5 - 20x^3 + 50x + 2) = 5x^4 - 60x^2 + 50$

Step 2: Find the critical points by setting the first derivative equal to zero.

We set $f'(x) = 0$ to find the critical points: $5x^4 - 60x^2 + 50 = 0$ Divide by 5: $x^4 - 12x^2 + 10 = 0$ Let $u = x^2$. Then the equation becomes: $u^2 - 12u + 10 = 0$ Using the quadratic formula to solve for $u$: $u = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(1)(10)}}{2(1)} = \frac{12 \pm \sqrt{144 - 40}}{2} = \frac{12 \pm \sqrt{104}}{2} = \frac{12 \pm 2\sqrt{26}}{2} = 6 \pm \sqrt{26}$ Since $u = x^2$, we have $x^2 = 6 \pm \sqrt{26}$.

Step 3: Solve for x.

Since $6 > \sqrt{26}$, both $6 + \sqrt{26}$ and $6 - \sqrt{26}$ are positive. Therefore, we can find the values of $x$: $x = \pm \sqrt{6 + \sqrt{26}}, \quad x = \pm \sqrt{6 - \sqrt{26}}$ This gives us four critical points: $x_1 = -\sqrt{6 + \sqrt{26}}, \quad x_2 = -\sqrt{6 - \sqrt{26}}, \quad x_3 = \sqrt{6 - \sqrt{26}}, \quad x_4 = \sqrt{6 + \sqrt{26}}$ Approximately, we have $x_1 \approx -3.2, x_2 \approx -1.0, x_3 \approx 1.0, x_4 \approx 3.2$

Step 4: Determine the sign of the function at intervals defined by critical points and large negative/positive values.

Let's analyze the sign of $f(x)$ at values around the critical points. We already know $f'(x) = 5(x^4 - 12x^2 + 10)$. We can check values for $f(x)$ at and around the critical points to determine the number of roots. Let $x_0$ be a very large negative number. Then $f(x_0)$ will be negative since the leading term is $x^5$. Let $x_1 = -\sqrt{6 + \sqrt{26}} \approx -3.2$. Then $f(x_1) \approx -54.7 - 20(-32.77) + 50(-3.2) + 2 \approx -54.7 + 655.4 - 160 + 2 \approx 442.7 > 0$. Let $x_2 = -\sqrt{6 - \sqrt{26}} \approx -1.0$. Then $f(x_2) \approx -1 - 20(-1) + 50(-1) + 2 = -1 + 20 - 50 + 2 = -29 < 0$. Let $x_3 = \sqrt{6 - \sqrt{26}} \approx 1.0$. Then $f(x_3) \approx 1 - 20 + 50 + 2 = 33 > 0$. Let $x_4 = \sqrt{6 + \sqrt{26}} \approx 3.2$. Then $f(x_4) \approx 335.5 - 20(32.77) + 50(3.2) + 2 \approx 335.5 - 655.4 + 160 + 2 \approx -157.9 < 0$. Let $x_5$ be a very large positive number. Then $f(x_5)$ will be positive since the leading term is $x^5$.

We have the following signs: $f(-\infty) < 0$ $f(x_1) > 0$ $f(x_2) < 0$ $f(x_3) > 0$ $f(x_4) < 0$ $f(\infty) > 0$

Since the sign changes 5 times, there are 5 roots.

Step 5: Conclude the number of roots.

The sign changes of $f(x)$ indicate the presence of roots in the intervals $(-\infty, x_1)$, $(x_1, x_2)$, $(x_2, x_3)$, $(x_3, x_4)$, and $(x_4, \infty)$. Therefore, there are 5 real roots.

Common Mistakes & Tips

• Sign Errors: Be very careful with signs when calculating the derivative and evaluating the function at critical points.
• Approximations: While approximations can help in understanding the relative positions of the critical points, avoid relying solely on approximations for determining the sign of the function, as this can lead to errors.
• Checking End Behavior: Always check the end behavior of the polynomial (as $x \rightarrow \pm \infty$) to ensure you haven't missed any roots.

Summary

To find the number of times the curve $y = x^5 - 20x^3 + 50x + 2$ crosses the x-axis, we found the critical points by setting the first derivative equal to zero. This gave us four critical points. We then analyzed the sign of the function in the intervals defined by these critical points, along with the end behavior of the function. The sign changes indicated the presence of 5 real roots.

The final answer is \boxed{5}.