Chapter 4.2, Problem 30E

To determine

To find: The critical number of the function $h\left(p\right)=\frac{p-1}{p^2+4}$.

Answer to Problem 30E

The critical number of the function $h\left(p\right)$ are $p=1+\sqrt{5}$ and $p=1+\sqrt{5}$.

Explanation of Solution

Definition used:

A critical number of a function f is a number c, if it satisfies either of the below conditions:

(1) $f^{\prime }\left(c\right)=0$

(2) $f^{\prime }\left(c\right)$ does not exist.

Quotient Rule:

If two functions $g\left(x\right)$ and $h\left(x\right)$ are differentiable and $h\left(x\right)\ne 0$, then the derivative of $f\left(x\right)=\frac{g\left(x\right)}{h\left(x\right)}$ is,

$f^{\prime }\left(x\right)=\frac{h\left(x\right)g^{\prime }\left(x\right)-g\left(x\right)h^{\prime }\left(x\right)}{{\left[h\left(x\right)\right]}^2}$ (1)

Calculation:

Obtain the first derivative of the given function.

$h^{\prime }\left(p\right)=\frac{d}{dp}\left(\frac{p-1}{p^2+4}\right)$

Apply the Quotient rule as shown in equation (1),

$\begin{array}{c}{h}^{\prime }\left(p\right)=\frac{\left(p^2+4\right)\left(1\right)-\left(p-1\right)\left(2p\right)}{{\left(p^2+4\right)}^2}\\ =\frac{p^2+4-2p^2+2p}{{\left(p^2+4\right)}^2}\\ =\frac{-p^2+2p+4}{{\left(p^2+4\right)}^2}\end{array}$

$h^{\prime }\left(p\right)=\frac{-\left(p^2-2p-4\right)}{{\left(p^2+4\right)}^2}$

Take $h^{\prime }\left(p\right)=0$ and obtain the critical numbers.

$\begin{array}{c}\frac{\left(p^2-2p-4\right)}{{\left(p^2+4\right)}^2}=0\\ p^2-2p-4=0\end{array}$

Obtain the roots of the equation by using the quadratic formula.

$\begin{array}{c}p=\frac{2\pm \sqrt{4+16}}{2}\\ =\frac{2\pm \sqrt{20}}{2}\\ =\frac{2\pm 2\sqrt{5}}{2}\\ =\frac{2\left(1\pm \sqrt{5}\right)}{2}\end{array}$

Therefore, the roots are, $p=1+\sqrt{5}$ and $p=1+\sqrt{5}$

Since the expression $p^2+4$ in the denominator of $h^{\prime }\left(p\right)$ never equal to zero, $h^{\prime }\left(p\right)$ exists for all real numbers.

Hence, it fails to satisfy the definition of critical number.

Thus, the critical numbers of the function $h\left(p\right)$ are $p=1+\sqrt{5}$ and $p=1+\sqrt{5}$.