To find the relative extreme points of the function $f(x) = (x - 5)^{1/3}$, we follow these steps:

Step 1: First Derivative

The relative extreme points occur where the derivative $f'(x)$ is zero or undefined. We start by finding the first derivative of $f(x)$.

Using the chain rule, we differentiate $f(x) = (x - 5)^{1/3}$:

$f'(x) = \frac{1}{3}(x - 5)^{-2/3} \cdot (1) = \frac{1}{3}(x - 5)^{-2/3}$

Thus, the derivative is:

$f'(x) = \frac{1}{3}(x - 5)^{-2/3}$

Step 2: Identify Critical Points

To find critical points, we solve $f'(x) = 0$ and check where the derivative is undefined.

1. Where is $f'(x) = 0$?

For $f'(x) = 0$, we set:

$\frac{1}{3}(x - 5)^{-2/3} = 0$

This equation has no solution because $(x - 5)^{-2/3}$ is never zero for any real value of $x$.

2. Where is $f'(x)$ undefined?

The derivative $f'(x)$ is undefined when $x = 5$, since the denominator involves $(x - 5)^{-2/3}$, which is undefined at $x = 5$.

Step 3: Second Derivative Test

Let's check if $x = 5$ is a potential extreme point by analyzing the behavior of the function around this point.

Second Derivative:

We find the second derivative $f''(x)$ by differentiating $f'(x)$:

$f'(x) = \frac{1}{3}(x - 5)^{-2/3}$

Differentiating this:

$f''(x) = -\frac{2}{9}(x - 5)^{-5/3}$

Thus,

$f''(x) = -\frac{2}{9}(x - 5)^{-5/3}$

The second derivative shows that the function is undefined at $x = 5$. To the left of $x = 5$, the second derivative is positive, and to the right, it is negative. This indicates that the function has a cusp at $x = 5$ and is not differentiable there.

Step 4: Conclusion

The function has no relative extreme points because the derivative does not become zero at any point. However, there is a cusp at $x = 5$, indicating a sharp change in the slope but no local maximum or minimum.

Step 5: Graph of the Function

The graph of $f(x) = (x - 5)^{1/3}$ looks like a cube root function, but it is shifted to the right by 5 units. The function is continuous and passes through $x = 5$ with a cusp. The slope approaches infinity near $x = 5$, and the function has no relative extrema.

Would you like a visual sketch of this graph?

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Here are some related questions to explore:

1. What is the significance of a cusp in a function's graph?
2. Can a function have a critical point where it is not differentiable?
3. What is the geometric meaning of a cube root function?
4. How does the chain rule apply to functions with fractional exponents?
5. Why does the derivative become undefined at certain points?

Tip: When a function has a cusp or corner at a point, it indicates that the function is continuous but not smooth at that point.