[ $0$ / $16$ . $66$ Points] $\newline$ DETAILS $\newline$ PREVIOUS ANSWERS $\newline$ SCALCET $8$ $4$ . $1$ . $036$ . $\newline$ Find the critical numbers of the function. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) $\newline$ $\newline$ $h(p)=\frac{p-4}{p^{2}+8}$ $\newline$ $p=\text{DNE}$ $\newline$ Submit Answer

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Algebra 2

Complex conjugate theorem

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Q. [

0

16

66

Points]

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DETAILS

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PREVIOUS ANSWERS

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SCALCET

8

4

1

036

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Find the critical numbers of the function. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.)

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h(p)=\frac{p-4}{p^{2}+8}

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p=\text{DNE}

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Submit Answer

Find Critical Numbers: To find the critical numbers of the function $h(p)$ , we need to find the values of $p$ where the derivative $h'(p)$ is either zero or undefined. $\newline$ First, let's find the derivative of $h(p)$ using the quotient rule, which states that if $f(p) = \frac{g(p)}{h(p)}$ , then $f'(p) = \frac{g'(p)h(p) - g(p)h'(p)}{(h(p))^2}$ . $\newline$ The derivative of the numerator $g(p) = p - 4$ is $g'(p) = 1$ . $\newline$ The derivative of the denominator $h(p) = p^2 + 8$ is $h'(p) = 2p$ . $\newline$ Now, we apply the quotient rule to find $h'(p)$ .
Derivative of h(p): Using the quotient rule, we have: $\newline$ h'(p) = $\frac{1 \cdot (p^2 + 8) - (p - 4) \cdot 2p}{(p^2 + 8)^2}$ $\newline$ Simplify the derivative: $\newline$ h'(p) = $\frac{p^2 + 8 - 2p^2 + 8p}{(p^2 + 8)^2}$ $\newline$ h'(p) = $\frac{-p^2 + 8p + 8}{(p^2 + 8)^2}$
Apply Quotient Rule: Now, we need to find the values of $p$ where $h'(p)$ is zero or undefined. $\newline$ The denominator $(p^2 + 8)^2$ is never zero because $p^2 + 8$ is always positive, so $h'(p)$ is never undefined. $\newline$ To find where $h'(p)$ is zero, we set the numerator equal to zero and solve for $p$ : $\newline$ $-p^2 + 8p + 8 = 0$
Simplify Derivative: We can solve the quadratic equation $-p^2 + 8p + 8 = 0$ by factoring or using the quadratic formula. However, this equation does not factor nicely, so we use the quadratic formula: $\newline$ $p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $\newline$ Here, $a = -1$ , $b = 8$ , and $c = 8$ .
Find Values of p: Plugging the values into the quadratic formula, we get: $\newline$ p = $\frac{-8 \pm \sqrt{8^2 - 4(-1)(8)}}{2(-1)}$ $\newline$ p = $\frac{-8 \pm \sqrt{64 + 32}}{-2}$ $\newline$ p = $\frac{-8 \pm \sqrt{96}}{-2}$
Quadratic Equation Solution: Simplify under the square root and the fraction: $\newline$ $p = \frac{-8 \pm \sqrt{16 \times 6}}{-2}$ $\newline$ $p = \frac{-8 \pm 4\sqrt{6}}{-2}$ $\newline$ $p = \frac{8 \mp 4\sqrt{6}}{2}$ $\newline$ $p = 4 \mp 2\sqrt{6}$
Simplify Quadratic Formula: We have two critical numbers from the quadratic equation: $\newline$ $p = 4 - 2\sqrt{6}$ and $p = 4 + 2\sqrt{6}$ $\newline$ These are the critical numbers where the derivative $h'(p)$ is zero.