What is the vertex of the function f(x) = x^3 + 8?

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Finding the Vertex of f(x) = x3 + 8

To determine the vertex of the function f(x) = x3 + 8, we first need to understand the nature of this function. In this case, we are dealing with a cubic function.

The general form of a cubic function is given by:

f(x) = ax3 + bx2 + cx + d

In our case, we can rewrite the function as:

f(x) = 1*x3 + 0*x2 + 0*x + 8

For cubic functions, the concept of a vertex, which is commonly associated with quadratic functions, does not apply in the same way. However, we can find critical points or inflection points that may provide insight into the function’s behavior.

Finding Critical Points

To find critical points, we need to calculate the derivative of the function:

f'(x) = 3x2

Setting the derivative equal to zero:

3x2 = 0

Solving for x gives:

x = 0

Now, we substitute this value back into the original function to find the corresponding y-coordinate:

f(0) = 03 + 8 = 8

Thus, we have a critical point at (0, 8).

Understanding the Function’s Behavior

Although (0, 8) can be considered a local extremum, it’s important to note that cubic functions do not have a vertex in the same sense as parabolas. The graph of this function will display an inflection at this critical point, which means it changes from concave down to concave up there.

In summary, while the function does not have a vertex, it does have a significant point at (0, 8) where its behavior changes. This point is indicative of the local extrema and the overall growth of the cubic function, providing a valuable reference for analyzing its characteristics.

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