Question

The range of the function f(k) = k2 + 2k + 1 is {25, 64}. What is the function’s domain? A. {5, 8} B. {-5, -8} C. {3, 8} D. {4, 7} E. {4, 8}

Answer 1

Answer:

Option D

The domain of the function is: {4, 7}

Step-by-step explanation:

We know that for polynomial functions like $f(k) = k^2 + 2k + 1$ 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

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

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

Now we factor the expression:

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

Then k = 4 and k = -6.

Now we do f(k) = 64

$k ^ 2 + 2k +1 = 64\\\\k ^ 2 + 2k -63 = 0$

We factor the expression:

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

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 2

Answer:

D. $[4,7]$

Step-by-step explanation:

The given function is

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

The  domain refers to the value of k, for which the function, $f(k)$ is defined.

Observe that the right hand side is a perfect square

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

We solve for k to get;

$k=\pm \sqrt{f(k)}-1$

when $f(k)=25$

$k=\pm \sqrt{25}-1$

$k=\pm5-1$

$k=4\:or\:k=-6$

when $f(k)=64$

$k=\pm \sqrt{64}-1$

$k=\pm8-1$

$k=7\:or\:k=-9$

The domain is $[4,7]$ or$[-9,-6]$

The correct answer is D.