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Beikeni · Answer 1 · 2023-08-10T05:15:42+0000

To answer this question, we need to evaluate the function:

Evaluating the function for x = √2

We can proceed as follows:

$k(\sqrt[]{2})=-(\sqrt[]{2})^2-2(\sqrt[]{2})+3=-(2^{(1)/(2)})^2-2\sqrt[]{2}+3$
$k(\sqrt[]{2})=-(2^{(2)/(2)})-2\sqrt[]{2}+3=-(2)-2\sqrt[]{2}+3=-2+3-2\sqrt[]{2}$
$k(\sqrt[]{2})=1-2\sqrt[]{2}$

We need to remember above that:

$\sqrt[]{n}=n^{(1)/(2)}$

Evaluating the function for x = a + 2

We can proceed as follows:

Then, we have:

Now, we need to add like terms as follows:

$k(a+2)=-a^2-6a-8+3\Rightarrow k(a+2)=-a^2-6a-5$

Therefore

Evaluating the function for x = -x

We have:

Evaluating the function for x = x²

We can proceed similarly:

Therefore, we have:

In summary, we have that:

$k(\sqrt[]{2})=1-2\sqrt[]{2}\approx-1.82842712475$
k(a+2)=-a^2-6a-5

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