Question

The range of the function f(k) = k2 + 2k + 1 is {25, 64}. What is the function’s domain?

Answer 1

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 2

Answer:

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

Step-by-step explanation:

$f(k) = k^2 + 2k + 1$

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

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

$25 = k^2 + 2k + 1$

Subtract 25 on both sides

$0= k^2 + 2k -24$

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

$0= (k+6)(k-4)$

$k+6=0, k=-6$

$k-4=0, k=4$

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

$64 = k^2 + 2k + 1$

Subtract 64  on both sides

$0= k^2 + 2k -63$

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

$0= (k+9)(k-7)$

$k+9=0, k=-9$

$k-7=0, k=7$

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

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