To find the critical points of the function $f(x) = \frac{1}{5}x^5 - \frac{8}{3}x^3 + 16x$, we need to follow these steps:

Step 1: Find the derivative $f'(x)$

The critical points occur where the derivative is equal to zero or undefined. Let's compute the derivative of $f(x)$:

$f'(x) = \frac{d}{dx} \left( \frac{1}{5}x^5 - \frac{8}{3}x^3 + 16x \right)$

Using the power rule for differentiation:

$f'(x) = \frac{1}{5} \cdot 5x^4 - \frac{8}{3} \cdot 3x^2 + 16$

Simplifying:

$f'(x) = x^4 - 8x^2 + 16$

Step 2: Set the derivative equal to zero

Now, we set $f'(x) = 0$ to find the critical points:

$x^4 - 8x^2 + 16 = 0$

This is a quadratic equation in terms of $x^2$. Let $u = x^2$, so the equation becomes:

$u^2 - 8u + 16 = 0$

Step 3: Solve the quadratic equation

Solve the quadratic equation $u^2 - 8u + 16 = 0$ using the quadratic formula:

$u = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(16)}}{2(1)}$

$u = \frac{8 \pm \sqrt{64 - 64}}{2}$

$u = \frac{8 \pm \sqrt{0}}{2}$

$u = \frac{8}{2} = 4$

Thus, $u = x^2 = 4$, so $x = \pm 2$.

Step 4: Find the critical points

The critical points are $x = 2$ and $x = -2$.

Conclusion:

The critical points of the function $f(x)$ are $x = 2$ and $x = -2$.

Would you like more details on analyzing the nature of these critical points (e.g., whether they are maxima, minima, or points of inflection)?

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Here are 5 questions that follow from this topic:

1. How do you determine if a critical point is a maximum or minimum?
2. What is the second derivative test, and how can it be applied to this problem?
3. How do inflection points differ from critical points in a function?
4. Can there be critical points where the derivative is undefined? If so, what types of functions might show this behavior?
5. How can graphing a function help in visualizing its critical points?

Tip: Always check the second derivative or use the first derivative test to confirm the nature of critical points (maxima, minima, or inflection points).