Problem 99 - Algebra for College Students

Question

Let \(f(x)=3 x-2, g(x)=-x^{2}+3 x-2,\) and \(h(x)=|x+2|\). Evaluate each expression. See Example 9. $$ \frac{g(2)}{h(-3)} $$

Step-by-Step Solution

Answer

0

Step 1 - Evaluate g(2)

First, substitute 2 into the function g(x). The function is given by \[ g(x) = -x^2 + 3x - 2 \]Substituting 2 for x, we get: \[ g(2) = -(2)^2 + 3(2) - 2 = -4 + 6 - 2 = 0 \]

Step 2 - Evaluate h(-3)

Next, substitute -3 into the function h(x). The function is given by \[ h(x) = |x + 2| \]Substituting -3 for x, we get: \[ h(-3) = |-3 + 2| = |-1| = 1 \]

Step 3 - Calculate g(2) / h(-3)

Now, divide the result of g(2) by the result of h(-3):\[ \frac{g(2)}{h(-3)} = \frac{0}{1} = 0 \]

Key Concepts

Functions

Functions are like machines that take an input and give an output based on a specific rule. Each function is usually written as \(f(x)\), where \(x\) is the input and \(f(x)\) is the output. For example, the function \(f(x) = 3x - 2\) takes any number \(x\), multiplies it by 3, and then subtracts 2 from the result. Similarly, \(g(x) = -x^2 + 3x - 2\) and \(h(x) = |x + 2|\) are rules for other functions. When you plug in a value for \(x\), you're 'evaluating' the function.

Absolute Value

The concept of absolute value is like taking the 'distance' from zero on a number line, without considering direction. The absolute value of a number \(x\) is written as \(|x|\). It turns negative numbers into positive ones. For example, \(|-3|\) is 3 because the distance from 0 to -3 is the same as the distance from 0 to 3. In the function \(h(x) = |x + 2|\), when you plug in a value of \(x\), you take the absolute value of \(x + 2\). So, if \(x = -3\), then \(h(-3) = |-3 + 2| = |-1| = 1\). Absolute value ensures your result is always non-negative.

Division

Division is one of the four basic arithmetic operations. It involves splitting a number into equal parts. When you divide one number by another, you're finding out how many times the divisor fits into the dividend. In fraction form, like \(\frac{a}{b}\), \(a\) is the dividend and \(b\) is the divisor. In this exercise, we needed to find the value of \(\frac{g(2)}{h(-3)}\). First, we evaluated \(g(2)\) and found it to be 0, then \(h(-3)\) as 1. The division \(\frac{0}{1}\) simplifies to 0 since zero divided by any non-zero number is always zero. Remember, dividing by zero is undefined.