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

Studdy Solution

STEP 1

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

STEP 2

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

STEP 3

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

STEP 4

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

STEP 5

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

STEP 6

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

STEP 7

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, $f(-6)$ is undefined.
So, $f(-6)$ is undefined is the correct answer for part (b).

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

Studdy Solution

STEP 1

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

STEP 2

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

STEP 3

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

STEP 4

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

STEP 5

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

STEP 6

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

STEP 7

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, $f(-6)$ is undefined.
So, $f(-6)$ is undefined is the correct answer for part (b).

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QuestionGiven the function f(x)=x−4x2−8f(x)=\frac{\sqrt{x-4}}{x^{2}-8}f(x)=x2−8x−4​​, find f(7)f(7)f(7) and f(−6)f(-6)f(−6). Are they defined?

Studdy Solution

STEP 1

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

STEP 2

First, let's evaluate f(7)f(7)f(7). We do this by substituting x=7x=7x=7 into the function.f(7)=7−472−8f(7)=\frac{\sqrt{7-4}}{7^{2}-8}f(7)=72−87−4​​

STEP 3

Now, simplify the expression inside the square root and the denominator.f(7)=349−8f(7)=\frac{\sqrt{3}}{49-8}f(7)=49−83​​

STEP 4

Calculate the value in the denominator.f(7)=341f(7)=\frac{\sqrt{3}}{41}f(7)=413​​So, f(7)=341f(7)=\frac{\sqrt{3}}{41}f(7)=413​​ is the correct answer for part (a).

STEP 5

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

STEP 6

Now, simplify the expression inside the square root and the denominator.f(−6)=−1036−8f(-6)=\frac{\sqrt{-10}}{36-8}f(−6)=36−8−10​​

STEP 7

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)f(−6) is undefined.So, f(−6)f(-6)f(−6) is undefined is the correct answer for part (b).

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

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

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

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

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

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

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

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, $f(-6)$ is undefined.
So, $f(-6)$ is undefined is the correct answer for part (b).