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Bismuth83
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For all real values of [m]x[/m], the function [m]f(x)[/m] has the form 26 Mar 2025, 21:23
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v: 0 w: \(\frac{-1}{4}\)
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For all real values of \(x\), the function \(f(x)\) has the form: \(f(x)=(x^2−4)^{−1}\). Select one value for \(v\) and one value for \(w\) that fulfills the condition that \(f(v) = w.\)
v w
\(\frac{-1}{4}\)
0
\(\frac{1}{4}\)
\(\frac{1}{3}\)
3
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v: 0 w: \(\frac{-1}{4}\)
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For all real values of [m]x[/m], the function [m]f(x)[/m] has the form 26 Mar 2025, 22:44
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For all real values of [m]x[/m], the function [m]f(x)[/m] has the form 26 Mar 2025, 23:45
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Bismuth83
For all real values of \(x\), the function \(f(x)\) has the form: \(f(x)=(x^2−4)^{−1}\). Select one value for \(v\) and one value for \(w\) that fulfills the condition that \(f(v) = w.\)
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Solution

\(f(x)=(x^2−4)^{−1}\) can be expressed as \(\frac{1}{(x+2)(x-2)}\). This tells us the value for x can be -2 or 2, and when assigning these values back to the equation, we will get two results,\(\frac{1}{(2+2)(2-2)}\)= 0 and \(\frac{1}{(-2+2)(-2-2)}\)=0. So, the value for v = 0 and from this if put 0 back to the given equation then we will get\( \frac{1}{(0^2-4)}\)=\( \frac{-1}{4}\) the value for w.


Bismuth83, is this approach correct?
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For all real values of [m]x[/m], the function [m]f(x)[/m] has the form 02 Jul 2025, 04:35
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No, you cant equate denominator to 0, it would make the function undefined. In the given function, x can take any values other than 2 and -2.
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Bismuth83
For all real values of \(x\), the function \(f(x)\) has the form: \(f(x)=(x^2−4)^{−1}\). Select one value for \(v\) and one value for \(w\) that fulfills the condition that \(f(v) = w.\)
Solution

\(f(x)=(x^2−4)^{−1}\) can be expressed as \(\frac{1}{(x+2)(x-2)}\). This tells us the value for x can be -2 or 2, and when assigning these values back to the equation, we will get two results,\(\frac{1}{(2+2)(2-2)}\)= 0 and \(\frac{1}{(-2+2)(-2-2)}\)=0. So, the value for v = 0 and from this if put 0 back to the given equation then we will get\( \frac{1}{(0^2-4)}\)=\( \frac{-1}{4}\) the value for w.


Bismuth83, is this approach correct?
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poojaarora1818
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For all real values of [m]x[/m], the function [m]f(x)[/m] has the form 03 Jul 2025, 12:07
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Thank you so much for your response. I appreciate it!
agrotechpepper
No, you cant equate denominator to 0, it would make the function undefined. In the given function, x can take any values other than 2 and -2.
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Bismuth83
For all real values of \(x\), the function \(f(x)\) has the form: \(f(x)=(x^2−4)^{−1}\). Select one value for \(v\) and one value for \(w\) that fulfills the condition that \(f(v) = w.\)
Solution

\(f(x)=(x^2−4)^{−1}\) can be expressed as \(\frac{1}{(x+2)(x-2)}\). This tells us the value for x can be -2 or 2, and when assigning these values back to the equation, we will get two results,\(\frac{1}{(2+2)(2-2)}\)= 0 and \(\frac{1}{(-2+2)(-2-2)}\)=0. So, the value for v = 0 and from this if put 0 back to the given equation then we will get\( \frac{1}{(0^2-4)}\)=\( \frac{-1}{4}\) the value for w.


Bismuth83, is this approach correct?
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svsivakrishna
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For all real values of [m]x[/m], the function [m]f(x)[/m] has the form 06 Jul 2025, 20:55
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Given f(x)=(x^2−4) ^ (−1)
To find: v and w such that
they fulfill the condition f(v)=w.

Therefore w=1/(v^2-4)

Considering from the answer choices, select v=0, then w= -1/4, both of which satisfy the given options.
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