Respuesta :

nicholaslisle
The domain of a function contains the x-values of the function while the range of the function contains the y-values of the function. In this case, we substitute 25 and 64 to y. Then,
25 = k2 + 2 k + 1 (k-4) * (k+6) = 0
64 = k2 + 2 k + 1 (k-7) * (k+9) = 0
hence the domain is {-9,-6, 4,7}

Answer:

Domain is [-9,-6] U [4,7]

Step-by-step explanation:

[tex]f(k) = k^2 + 2k + 1[/tex]

Range is {25, 64}. Range is the y value

LEts plug in 25 for f(k) and solve for k

[tex]25 = k^2 + 2k + 1[/tex]

Subtract 25 on both sides

[tex]0= k^2 + 2k -24[/tex]

Product is -24 and sum is +2. factors are 6 and -4

[tex]0= (k+6)(k-4)[/tex]

[tex]k+6=0, k=-6[/tex]

[tex]k-4=0, k=4[/tex]

Do the same and solve for x when f(x)= 64

[tex]64 = k^2 + 2k + 1[/tex]

Subtract 64  on both sides

[tex]0= k^2 + 2k -63[/tex]

Product is -63 and sum is +2. factors are 9 and -7

[tex]0= (k+9)(k-7)[/tex]

[tex]k+9=0, k=-9[/tex]

[tex]k-7=0, k=7[/tex]

We got k values, -9,-6,4,7

Domain is [-9,-6] U [4,7]

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