Problem 15 - Precalculus

Question

For the functions \(f(x)=3^{x}, g(x)=\left(\frac{1}{16}\right)^{x},\) and \(h(x)=10^{x+1},\) find the function value at the indicated points. $$g(-1)$$

Step-by-Step Solution

Answer

The value of \( g(-1) \) is 16.

Step 1: Understand the Function

We are given the function \( g(x) = \left(\frac{1}{16}\right)^x \). This is an exponential function where the base is \( \frac{1}{16} \). We need to evaluate this function at \( x = -1 \).

Step 2: Substitute -1 for x

Substitute \(-1\) in place of \(x\) in the function \( g(x) \), so it becomes \( g(-1) = \left(\frac{1}{16}\right)^{-1} \).

Step 3: Apply the Negative Exponent Rule

Recall that a negative exponent means the reciprocal of the base raised to the positive of that exponent. Therefore, \( \left(\frac{1}{16}\right)^{-1} = 16 \).

Step 4: Evaluate the Expression

Using the calculations from the previous step, we find \( g(-1) = 16 \).

Key Concepts

Negative Exponent Rule

Understanding exponents is key to grasping many mathematical concepts. The Negative Exponent Rule tells us how to handle exponents that are less than zero. When you see a negative exponent, such as in the expression \( a^{-n} \), it means we're looking at the reciprocal of the base \( a \) raised to the positive power \( n \).
For example, \( a^{-n} = \frac{1}{a^n} \). This method turns what could be complex exponential expressions into more manageable fractions.

• Negative exponents shrink numbers less than 1.
• They "flip" larger numbers into fractions.
• This rule is universally applicable for both whole numbers and fractions as the base.

Let’s consider \(\left(\frac{1}{16}\right)^{-1}\). According to the rule, it becomes \(16^1\), which simplifies to 16. Understanding this concept not only simplifies calculations but helps when working with powers and roots.

Function Evaluation

Exponential functions, such as \(g(x) = \left(\frac{1}{16}\right)^x\), are an essential element of algebra and calculus. Evaluating a function simply means to find its output for a given input. Here we focus on finding the value of \(g(x)\) when \(x = -1\).
To evaluate, substitute the given value directly into the function where the variable \(x\) appears. So, for \(g(-1)\), it becomes:

• Replace \(x\) in the function with \(-1\).
• It turns into \(\left(\frac{1}{16}\right)^{-1}\).
• Now, apply the relevant exponential rules.

Through realistic examples like this, students develop an intuitive understanding. The process doesn't change for different values of \(x\); substituting and applying any necessary rules will always guide you to the correct evaluation.

Reciprocal

In mathematics, understanding the concept of reciprocal is crucial for mastering negative exponentials and divisions. The reciprocal of a number \(a\) is \(\frac{1}{a}\). This flips the number over and swaps the numerator with the denominator.
Reciprocals are especially important when dealing with negative exponents because they often convert the expressions and simplify our calculations.

• The reciprocal of \(16\) is \(\frac{1}{16}\).
• Similarly, \(\left(\frac{1}{16}\right)\) becomes \(16\) when considering the negative exponent \(-1\).
• Reciprocals turn equations and expressions manageable by simplifying complex fractions.

With terminology like 'flip' for reciprocals, we can ground mathematical theory in tangible imagery, making calculations less intimidating and more engaging to solve.