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QuestionUse the function f(x)=x−5f(x)=\sqrt{x-5} to find f(x)f(x) for x=1,5,86,105x=1, 5, 86, 105. Simplify where possible.

Studdy Solution

STEP 1

Assumptions1. The function rule is f(x)=x−5f(x)=\sqrt{x-5} . The values of xx we need to evaluate are1,5,86, and1053. If the value under the square root is negative, the result is not a real number

STEP 2

We need to substitute the values of xx into the function rule to find the corresponding f(x)f(x) values.

STEP 3

First, let's substitute x=1x=1 into the function rule.
f(1)=1−5f(1)=\sqrt{1-5}

STEP 4

implify the expression under the square root.
f(1)=−4f(1)=\sqrt{-4}

STEP 5

Since the value under the square root is negative, f(1)f(1) is not a real number.

STEP 6

Next, let's substitute x=5x=5 into the function rule.
f(5)=5−5f(5)=\sqrt{5-5}

STEP 7

implify the expression under the square root.
f(5)=0f(5)=\sqrt{0}

STEP 8

Calculate the square root of0.
f(5)=0f(5)=0

STEP 9

Next, let's substitute x=86x=86 into the function rule.
f(86)=86−5f(86)=\sqrt{86-5}

STEP 10

implify the expression under the square root.
f(86)=81f(86)=\sqrt{81}

STEP 11

Calculate the square root of81.
f(86)=9f(86)=9

STEP 12

Finally, let's substitute x=105x=105 into the function rule.
f(105)=105−5f(105)=\sqrt{105-5}

STEP 13

implify the expression under the square root.
f(105)=100f(105)=\sqrt{100}

STEP 14

Calculate the square root of100.
f(105)=10f(105)=10

STEP 15

Now we can fill in the table with the calculated f(x)f(x) values.
\begin{tabular}{|c|c|} \hlinexx & f(x)f(x) \\ \hline & Not a real number \\ \hline5 &0 \\ \hline86 &9 \\ \hline105 &10 \\ \hline\end{tabular}

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