Problem 54 - College Algebra with Modeling and Visualization

Question

Exercises \(53-56:\) Use the given \(f(x)\) and \(g(x)\) to evaluate each expression. $$ \begin{aligned} &f(x)=\left|x^{2}-4\right|, \quad g(x)=2 x^{2}+x+1\\\ &\begin{array}{lll} \text { (a) }(f \circ g)(1) & \text { (b) }(g \circ f)(-3) \end{array} \end{aligned} $$

Step-by-Step Solution

Answer

(a) 12, (b) 56

Step 1: Understand the Function Notation

In function notation, \((f \circ g)(x)\) represents the composition of functions, meaning \(f(g(x))\). Similarly, \((g \circ f)(x)\) means \(g(f(x))\). Our task is to find these compositions evaluated at given points.

Step 2: Evaluate \(g(1)\)

Substitute \(x = 1\) into the function \(g(x) = 2x^2 + x + 1\).\[g(1) = 2(1)^2 + 1 + 1 = 2 + 1 + 1 = 4\]

Step 3: Evaluate \(f(g(1))\)

Now we use the value obtained in the previous step to find \(f(g(1))\), which is \(f(4)\).\[f(x) = |x^2 - 4|\] Substitute \(x = 4\):\[f(4) = |4^2 - 4| = |16 - 4| = |12| = 12\]

Step 4: Evaluate \(f(-3)\)

Substitute \(x = -3\) into the function \(f(x) = |x^2 - 4|\).\[f(-3) = |(-3)^2 - 4| = |9 - 4| = |5| = 5\]

Step 5: Evaluate \(g(f(-3))\)

Now that we have \(f(-3) = 5\), substitute this into \(g(x) = 2x^2 + x + 1\) to find \(g(f(-3))\) or \(g(5)\).\[g(5) = 2(5)^2 + 5 + 1 = 2(25) + 5 + 1 = 50 + 5 + 1 = 56\]

Step 6: Final Solutions to Expressions

We have calculated:- \((f \circ g)(1) = 12\)- \((g \circ f)(-3) = 56\)

Key Concepts

Function Notation

Function notation is a straightforward way to express functions, making them easier to work with. When you see something like \( f(x) \), it means we have a function \( f \) that depends on the variable \( x \). The letter inside the parentheses \( x \) is called the input, which is used to find the output by performing the operations defined in the function. So, \( f(x) \) gives us the result after applying the function's rule to the input \( x \).

In the concept of function composition, such as \( (f \circ g)(x) \), it indicates a combination of two functions. It means you first apply \( g(x) \) and then apply \( f(x) \) to the result. This is read as 'f composed with g of x.' This helps in creating more complex functions from simpler ones.

Keep in mind:

• \( (f \circ g)(x) = f(g(x)) \)
• \( (g \circ f)(x) = g(f(x)) \)

Absolute Value Function

An absolute value function is characterized by the symbol \( | \, | \). It measures the distance of a number from zero on the number line. No matter whether the input is positive or negative, the absolute value is always non-negative.

Consider a simple example: the absolute value of \(-3\) is 3, shown as \(|-3| = 3\). For the function \( f(x) = |x^2 - 4| \), the absolute value ensures the result is always positive or zero, turning any negative output of \( x^2 - 4 \) into a positive one.

Some key points about absolute value functions include:

• Always returns a non-negative value.
• Reflects the input expression across the x-axis if it dips below zero.
• Is useful in distance and magnitude calculations.

Quadratic Function

Quadratic functions are a type of polynomial function where the highest degree of the variable is squared (i.e. \( x^2 \)). The standard form of a quadratic function is \( ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \).

In our solution, the quadratic function is \( g(x) = 2x^2 + x + 1 \). It forms a parabola when graphed, opening upwards if \( a > 0 \) or downwards if \( a < 0 \). The graph's vertex is the highest or lowest point, and the axis of symmetry runs through this vertex, demonstrated via the formula \( x = -\frac{b}{2a} \).

Points to remember about quadratic functions:

• They can model real-world situations such as projectile motion.
• The vertex form \( a(x-h)^2 + k \) highlights the vertex \((h, k)\) directly.
• They always produce a symmetrical graph about the vertical axis passing through the vertex.