Problem 45 - College Algebra with Corequisite Support

Question

For the following exercises, evaluate the exponential functions for the indicated value of \(x\). \(h(x)=-\frac{1}{2}\left(\frac{1}{2}\right)^{x}+6\) for $$ h(-7) $$

Step-by-Step Solution

Answer

The value of \( h(-7) \) is \(-58\).

Step 1: Substitute the Value into the Function

We begin by substituting the value of \(x\) into the function \(h(x)\). For \(h(-7)\), replace \(x\) with \(-7\): \[ h(-7) = -\frac{1}{2}\left(\frac{1}{2}\right)^{-7} + 6 \]

Step 2: Apply the Exponent Rule

Applying the exponent rule for negative exponents, which states \(a^{-b} = \frac{1}{a^b}\), let's evaluate \(\left(\frac{1}{2}\right)^{-7}\):\[ \left(\frac{1}{2}\right)^{-7} = 2^7 \]

Step 3: Calculate the Power

Now, compute \(2^7\): \[ 2^7 = 128 \]

Step 4: Calculate and Simplify the Expression

Substitute back the value of \(2^7\) into the expression: \[ h(-7) = -\frac{1}{2} \times 128 + 6 \]Simplify the multiplication:\[ h(-7) = -64 + 6 \]

Step 5: Final Calculation

Complete the calculation by adding \(-64\) to \(6\): \[ h(-7) = -58 \]

Key Concepts

Understanding Negative Exponents

Negative exponents can be a bit confusing at first, but they follow a very straightforward rule. When you have a number with a negative exponent, it means you take the reciprocal of the base and then apply the positive version of the exponent. In other words:

• For an expression like \( a^{-b} \), you would think of it as \( \frac{1}{a^b} \). This turns a negative exponent into something manageable by flipping the base over and exponentiating it positively.
• In our example with \( \left(\frac{1}{2}\right)^{-7} \), you flip \( \frac{1}{2} \) to become \( 2 \) raised to the power of 7: \( 2^7 \).

Remember, the negative sign does not make the base negative but instead inverts it into a fraction if it's not already one. This tidbit is crucial when you evaluate expressions with negative exponents.

The Substitution Method Explained

The substitution method is essential when you want to find the value of a function for a specific variable. The idea here is simple: you have a function, and you're given a specific value to plug into it. Here's how it works:

• Identify the given function and the value for which you need to evaluate it. In the problem, we have \( h(x) = -\frac{1}{2}\left(\frac{1}{2}\right)^x + 6 \) and we need to evaluate \( h(-7) \).
• Next, plug in the value of \( x \) into the function. Replace every instance of \( x \) with \(-7\). Your expression then starts looking like this: \( h(-7) = -\frac{1}{2}\left(\frac{1}{2}\right)^{-7} + 6 \).

The substitution method is a step-by-step way to simplify a problem by transforming it from a more generalized expression to a specific numerical one. This method is fundamental in both algebra and calculus.

Evaluating Functions Step-by-Step

Evaluating a function means finding the output value for a given input. It’s straightforward when you follow the steps, but here's how to make sense of it:

• First, make sure you've substituted correctly into the function. This ensures you're working with the right numbers.
• Next, deal with any exponents, especially negative ones, using the rules we've discussed. In our case, we've dealt with \( \left(\frac{1}{2}\right)^{-7} \) by calculating it as \( 2^7 = 128 \).
• Afterwards, carry out any multiplication or division as necessary. For instance, here \( -\frac{1}{2} \times 128 = -64 \).
• Finally, add or subtract remaining terms to reach the solution, just like adding \( 6 \) to \(-64 \) to get the final answer, \(-58\).

The focus when evaluating any function is to simplify and solve efficiently, ensuring each operation builds towards an accurate final answer.