Problem 66 - Intermediate Algebra

Question

Find each function value, if possible. Do not use a calculator. See Example 5. $$ s(a)=-\sqrt[3]{32 a} $$ a. \(s(-2)\) b. \(s(2)\)

Step-by-Step Solution

Answer

s(-2) = 4; s(2) = -4.

Step 1: Substitute the Value into the Function for s(-2)

To find \(s(-2)\), substitute \(a = -2\) into the function. The function is \(s(a) = -\sqrt[3]{32a}\). Therefore, \(s(-2) = -\sqrt[3]{32 \times (-2)}\).

Step 2: Calculate Inside the Cube Root for s(-2)

Calculate the expression inside the cube root: \(32 \times (-2) = -64\).

Step 3: Evaluate the Cube Root for s(-2)

Calculate the cube root of \(-64\): \(\sqrt[3]{-64} = -4\), since \((-4)^3=-64\).

Step 4: Apply the Negative Sign for s(-2)

Apply the negative sign from the original function: \(s(-2) = -(-4) = 4\).

Step 5: Substitute the Value into the Function for s(2)

To find \(s(2)\), substitute \(a = 2\) into the function. The function is \(s(a) = -\sqrt[3]{32a}\). Therefore, \(s(2) = -\sqrt[3]{32 \times 2}\).

Step 6: Calculate Inside the Cube Root for s(2)

Calculate the expression inside the cube root: \(32 \times 2 = 64\).

Step 7: Evaluate the Cube Root for s(2)

Calculate the cube root of \(64\): \(\sqrt[3]{64} = 4\), since \(4^3 = 64\).

Step 8: Apply the Negative Sign for s(2)

Apply the negative sign from the original function: \(s(2) = -4\).

Key Concepts

Substitution

When it comes to functions, substitution is like replacing a part of a recipe with a specific ingredient. You take the given values of variables and plug them into the function. This helps you find the function's output for those specific values. In our case, we have the function \( s(a) = -\sqrt[3]{32a} \).

Let's look at how substitution works:

• First, identify the function and the value of \(a\) you need to substitute. For part (a), we use \(a = -2\), and for part (b), \(a = 2\).
• Next, in \(a = -2\), switch out every "\(a\)" in the function with \(-2\), leading to \( s(-2) = -\sqrt[3]{32 \times (-2)}\).
• Do the same for \(a = 2\), resulting in \( s(2) = -\sqrt[3]{32 \times 2}\).

Through substitution, you're preparing to calculate the final value of the function for the given inputs.

Cube Root

The cube root is about finding a number that, when multiplied by itself three times, gives you the original number under the root. In mathematical terms, if \(x^3 = y\), then \(x = \sqrt[3]{y}\). The operation is similar to finding a square root, but instead, we're working in threes.

Here's how it applies to our function:

• For \(s(-2)\), solve the cube root of \(-64\), since \(32 \times (-2) = -64\). We know \(-4\) cubed equals \(-64\), so \(\sqrt[3]{-64} = -4\).
• For \(s(2)\), solve the cube root of \(64\), because \(32 \times 2 = 64\). Here, \(4^3 = 64\), thus \(\sqrt[3]{64} = 4\).

Understanding cube roots is crucial as it allows us to simplify parts of the function and find solutions without needing a calculator.

Negative Sign

A negative sign can change the whole value of an expression and is essential to understand when tackling function problems. The minus sign in front of a function or a constant indicates that the result will be inverse in sign, compared to what you calculate internally.

Let's see how it affects our function outcomes:

• Once you find the cube root for \(s(-2)\) as \(-4\), applying the negative sign from the function results in \(-(-4)\). Double negatives turn into a positive, resulting in \(s(-2) = 4\).
• For \(s(2)\), after calculating the cube root as \(4\), applying the function's negative sign changes the result to \(-4\), making \(s(2) = -4\).

Handling negative signs correctly ensures that you derive the correct function values, adding accuracy to your calculations.