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Math

Math Snap

PROBLEM

Given the function f(x)=x4x28f(x)=\frac{\sqrt{x-4}}{x^{2}-8}, find f(7)f(7) and f(6)f(-6). Are they defined?

STEP 1

Assumptions1. The function is given by f(x)=x4x8f(x)=\frac{\sqrt{x-4}}{x^{}-8}
. We need to evaluate the function at x=7x=7 and x=6x=-6

STEP 2

First, let's evaluate f(7)f(7). We do this by substituting x=7x=7 into the function.
f(7)=74728f(7)=\frac{\sqrt{7-4}}{7^{2}-8}

STEP 3

Now, simplify the expression inside the square root and the denominator.
f(7)=3498f(7)=\frac{\sqrt{3}}{49-8}

STEP 4

Calculate the value in the denominator.
f(7)=341f(7)=\frac{\sqrt{3}}{41}So, f(7)=341f(7)=\frac{\sqrt{3}}{41} is the correct answer for part (a).

STEP 5

Now, let's evaluate f()f(-). We do this by substituting x=x=- into the function.
f()=4()28f(-)=\frac{\sqrt{--4}}{(-)^{2}-8}

STEP 6

Now, simplify the expression inside the square root and the denominator.
f(6)=10368f(-6)=\frac{\sqrt{-10}}{36-8}

SOLUTION

Here, we notice that the value inside the square root is negative. In real numbers, the square root of a negative number is undefined. Therefore, f(6)f(-6) is undefined.
So, f(6)f(-6) is undefined is the correct answer for part (b).

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