Answer:
Option D
The domain of the function is: {4, 7}
Step-by-step explanation:
We know that for polynomial functions like [tex]f(k) = k^2 + 2k + 1[/tex] its domain and its rank are all real numbers. However, for this case we are told that the function range is: the set {25, 64}
This means that the function is bounded.
Then the domain of f(k) are all possible values of k such that f(k) belongs to the interval {25, 64}.
To find the limit values of k then we do f(k) = 25
[tex]k^2 + 2k + 1 = 25[/tex]
[tex]k^2 + 2k -24 = 0[/tex]
Now we factor the expression:
[tex](k + 6)(k-4) = 0[/tex]
Then k = 4 and k = -6.
Now we do f(k) = 64
[tex]k ^ 2 + 2k +1 = 64\\\\k ^ 2 + 2k -63 = 0[/tex]
We factor the expression:
[tex](k + 9)(k-7) = 0[/tex]
k = -9 and k = 7.
Finally we search between the options given an interval that matches.
The option that matches is option D {4, 7}
Answer:
D. [tex][4,7][/tex]
Step-by-step explanation:
The given function is
[tex]f(k)=k^2+2k+1[/tex]
The domain refers to the value of k, for which the function, [tex]f(k)[/tex] is defined.
Observe that the right hand side is a perfect square
[tex]f(k)=(k+1)^2[/tex]
We solve for k to get;
[tex]k=\pm \sqrt{f(k)}-1[/tex]
when [tex]f(k)=25[/tex]
[tex]k=\pm \sqrt{25}-1[/tex]
[tex]k=\pm5-1[/tex]
[tex]k=4\:or\:k=-6[/tex]
when [tex]f(k)=64[/tex]
[tex]k=\pm \sqrt{64}-1[/tex]
[tex]k=\pm8-1[/tex]
[tex]k=7\:or\:k=-9[/tex]
The domain is [tex][4,7][/tex] or[tex][-9,-6][/tex]
The correct answer is D.
