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Chapter 1: Problem 34

Evaluate each function at the given values of the independent variable and simplify. \(f(r)=\sqrt{25-r}-6\) a. \(f(16)\) b. \(f(-24)\) c. \(f(25-2 x)\)

Short Answer

Expert verified
After evaluating the function at the given values and simplifying, we have \(f(16) = -3\), \(f(-24) = 1\), and \(f(25-2x) = \sqrt{2x} - 6\).

Step by step solution

01

Substitute r with 16

First we need to substitute \(r\) in our function with 16. When we do this, we get \(f(16) = \sqrt{25 - 16} - 6\). It's important to solve the operation within the square root first.
02

Simplify the operation

The operation inside the square root is \(25 - 16\) which results in 9. After substituting this result back into the function we get \(f(16) = \sqrt{9} - 6\). Taking the square root of 9 gives us 3. The function then simplifies to \(f(16) = 3 - 6\), which finally simplifies to \(f(16) = -3\).
03

Substitute r with -24

Next, repeat the same steps for \(r = -24\). Substitute \(-24\) for \(r\) in our function. This gives \(f(-24) = \sqrt{25 - (-24)} - 6\). This simplifies to \(f(-24) = \sqrt{49} - 6\).
04

Simplify the operation

\(\sqrt{49}\) equals 7. The function simplifies to \(f(-24) = 7 - 6\), which finally simplifies to \(f(-24) = 1\).
05

Substitute r with \(25-2x\)

Now for \(r = 25 - 2x\). Substituting \(25 - 2x\) for \(r\) gives \(f(25-2x) = \sqrt{25 - (25 - 2x)} - 6\). Simplifying this expression inside the square root gives \(f(25-2x) = \sqrt{2x} - 6\). This is the function \(f(r)\) evaluated at \(r = 25 - 2x\).

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