Problem 74 - College Algebra

Question

Use each pair of functions to find \(f(g(0))\) and \(g(f(0))\). $$ f(x)=\sqrt{x+4}, g(x)=12-x^{3} $$

Step-by-Step Solution

Answer

Both \(f(g(0))\) and \(g(f(0))\) equal 4.

Step 1: Evaluate g(0)

To find \(f(g(0))\), we first need to evaluate \(g(0)\). Substitute \(x = 0\) into the function \(g(x) = 12 - x^3\).\[g(0) = 12 - 0^3 = 12\]

Step 2: Evaluate f(g(0))

Now that we have \(g(0) = 12\), substitute this result into the function \(f(x) = \sqrt{x+4}\):\[f(g(0)) = f(12) = \sqrt{12 + 4} = \sqrt{16} = 4\]

Step 3: Evaluate f(0)

To find \(g(f(0))\), first evaluate \(f(0)\) by substituting \(x = 0\) into \(f(x) = \sqrt{x+4}\):\[f(0) = \sqrt{0 + 4} = \sqrt{4} = 2\]

Step 4: Evaluate g(f(0))

Now substitute the result \(f(0) = 2\) into the function \(g(x) = 12 - x^3\):\[g(f(0)) = g(2) = 12 - 2^3 = 12 - 8 = 4\]

Key Concepts

Function Evaluation

Function evaluation is essentially about plugging a number into a function and calculating it. This is like substituting ingredients into a recipe to see the final dish.
To evaluate a function, you need a clear understanding of both the function itself and the values you're dealing with.

• For example, if you have a function such as \( g(x) = 12 - x^3 \), and you're asked to find \( g(0) \), you simply replace every \( x \) with \( 0 \).
• This gives us \( g(0) = 12 - 0^3 = 12 \).

When dealing with composite functions like \( f(g(0)) \), you approach it layer by layer:
1. First, find \( g(0) \).
2. Then, take this result and substitute it into \( f(x) \).
Remember, carefully substituting and calculating each step will eventually lead to the correct answer.

Square Root Functions

Square root functions can seem tricky, but they're quite straightforward once you get the hang of them. The basic form of a square root function is \( f(x) = \sqrt{x} \).
This function takes the square root of its input, \( x \). When evaluating, it's important to remember that the square root of a number is a value that, when multiplied by itself, gives the original number.

• For instance, \( f(x) = \sqrt{x + 4} \) requires you to add 4 to \( x \) before applying the square root.
• If the input is 12, then \( f(12) = \sqrt{12 + 4} = \sqrt{16} = 4 \).

Square roots often connect to real-world problems involving areas and dimensions, where square root operations help determine lengths or rates.
Just ensure you perform operations inside the function's parentheses first, keeping everything straightforward and accurate.

Cubic Functions

Cubic functions involve the cube (or third power) of their input. A typical cubic function like \( g(x) = 12 - x^3 \) is always important to manage carefully because they feature nonlinear change.
These functions aren't just simple repeats like linear or quadratic functions. Instead, they change at a rate that's neither constant nor simple to predict just by looking.

• If you want to evaluate \( g(x) \) when \( x=0 \), calculating \( 0^3 \) is straightforward since it's simply \( 0 \).
• On the other hand, \( g(2) = 12 - (2)^3 = 12 - 8 = 4 \)

This takes checking or double-checking your arithmetic since small mistakes in power evaluation may lead to misinterpretation of the function's overall behavior.
Cubic functions often reflect in various studies from physics to engineering, where understanding growth, decay, or vibration speed might come in handy.