91Ó°ÊÓ

The range of a function includes all possible output values that the function can produce. For the cube root function \( f(x) = \sqrt[3]{8x - 24} \), determining the range is straightforward. Unlike square root functions, cube root functions are defined for all real inputs without restrictions, because they can take both positive and negative values inside the root.
What does this mean practically? It means that no matter the value of \(x\), \( f(x) \) will always produce a real number. This makes the function’s range very broad:

• The range spans all real numbers, \((-\infty, \infty)\).

Cube root functions don’t hit a ceiling or a floor in values they can produce. They closely resemble linear functions, due to their unrestricted nature.
When graphing \( f(x) = \sqrt[3]{8x - 24} \), you can observe that the curve extends infinitely up and down, confirming the function’s unlimited range.

The concept of increasing intervals refers to sections of a graph where the output values steadily rise as the input values increase. Understanding whether a function is increasing or not relies heavily on its derivative. For the given function:

• The derivative is \( f'(x) = \frac{8}{3(8x-24)^{2/3}} \).

Since cube root functions naturally exhibit a consistent upward slope, their derivative (\( f'(x) \)) remains positive over the entire domain.
This positive derivative indicates that \( f(x) \) is increasing everywhere it’s defined. Consequently:

• The largest interval where \( f(x) \) is increasing is precisely its entire domain, and that is all real numbers.

To clarify further, for any two numbers \( a \) and \( b \) where \( a < b \), \( f(a) < f(b) \), reinforcing that the function continually rises.

Sometimes, functions have scenes on their graphs where the output lessens as the input grows, which are called "decreasing intervals." However, not every function will have these intervals. In the case of our cube root function \( f(x) = \sqrt[3]{8x - 24} \), its behavior is exclusively increasing.
Why doesn't it ever decrease? This ties back to its derivative:

• With \( f'(x) \) being positive across all inputs, there are no spots where the function’s rate of change dips below zero.

Since \( f(x) \) constantly steps upwards, there isn’t any interval where it descends.
Thus, we can confidently assert:

• There is no interval on this graph where the function is decreasing.

Understanding this aspect underscores the function's tendency to climb regardless of input, emphasizing its steady increase across its entire domain.

When facing equations like \( f(x) = 0 \), solving them graphically means finding where the curve of the function touches the x-axis. This point where the function hits the x-axis is known as the "zero" or "root" of the function.
For \( f(x) = \sqrt[3]{8x - 24} \), we need to determine when the output equals zero, which is the x-intercept. To do this:

• Solve the equation \( 8x - 24 = 0 \).
• This simplifies to \( x = 3 \).

Plotting this on a graph confirms that at \( x = 3 \), \( f(x) \) touches the x-axis. This visual representation helps verify our algebraic solution.
With graphical solutions, it becomes easier to see where turning points and intercepts occur, offering a clear picture of the function’s behavior. It’s a handy tool, especially when algebraic solutions are tedious or when verifying accuracy in a quick, reliable manner.