Question

How do you find the critical numbers for $\dfrac{2-x}{{{\left( x+2 \right)}^{3}}}$ to determine the maximum and minimum?

Answer 1

Hint: In order to find the critical numbers for the given function we will first calculate the value of first order derivative of the given function i.e. $\dfrac{dy}{dx}$ if the given function is $y=\dfrac{2-x}{{{\left( x+2 \right)}^{3}}}$. Then we will equate the value of $\dfrac{dy}{dx}$ to zero to find the critical points.

Complete step by step solution:
We have been given an expression $\dfrac{2-x}{{{\left( x+2 \right)}^{3}}}$.
We have to find the critical numbers for the given expression.
Let us assume that the given function is $y=\dfrac{2-x}{{{\left( x+2 \right)}^{3}}}$.
To find the critical points first we have to find the value of $\dfrac{dy}{dx}$.
Now, differentiating the given function with respect to x we will get
$\Rightarrow y=\dfrac{d}{dx}\left( \dfrac{2-x}{{{\left( x+2 \right)}^{3}}} \right)$
Now, we know that by quotient rule of differentiation $\dfrac{d}{dx}\left( \dfrac{f(x)}{g(x)} \right)=\dfrac{g(x)\dfrac{d}{dx}f(x)-f(x)\dfrac{d}{dx}g(x)}{{{\left( g(x) \right)}^{2}}}$
Here we have $f(x)=\left( 2-x \right)$ and $g(x)={{\left( x+2 \right)}^{3}}$
Now, applying the formula to the given function we will get
\[\Rightarrow \dfrac{dy}{dx}=\left( \dfrac{{{\left( x+2 \right)}^{3}}\dfrac{d}{dx}\left( 2-x \right)-\left( 2-x \right)\dfrac{d}{dx}{{\left( x+2 \right)}^{3}}}{{{\left( {{\left( x+2 \right)}^{3}} \right)}^{2}}} \right)\]
Now, we know that $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$
Now, applying the formula and simplifying the above obtained equation we will get
\[\begin{align}
  & \Rightarrow \dfrac{dy}{dx}=\left( \dfrac{{{\left( x+2 \right)}^{3}}\left( -1 \right)-\left[ \left( 2-x \right)3{{\left( x+2 \right)}^{2}} \right]}{{{\left( x+2 \right)}^{6}}} \right) \\
 & \Rightarrow \dfrac{dy}{dx}=\left( \dfrac{-{{\left( x+2 \right)}^{3}}-\left[ \left( 2-x \right)3{{\left( x+2 \right)}^{2}} \right]}{{{\left( x+2 \right)}^{6}}} \right) \\
\end{align}\]
Now, putting the value of $\dfrac{dy}{dx}$ equal to zero we will get
\[\Rightarrow \left( \dfrac{-{{\left( x+2 \right)}^{3}}-\left[ \left( 2-x \right)3{{\left( x+2 \right)}^{2}} \right]}{{{\left( x+2 \right)}^{6}}} \right)=0\]
Now, simplifying the above obtained equation we will get
\[\begin{align}
  & \Rightarrow -{{\left( x+2 \right)}^{3}}-\left[ \left( 2-x \right)3{{\left( x+2 \right)}^{2}} \right]=0 \\
 & \Rightarrow -{{\left( x+2 \right)}^{3}}=3{{\left( x+2 \right)}^{2}}\left( 2-x \right) \\
 & \Rightarrow -\left( x+2 \right)=3\left( 2-x \right) \\
 & \Rightarrow -x-2=6-3x \\
 & \Rightarrow -x+3x=6+2 \\
 & \Rightarrow 2x=8 \\
 & \Rightarrow x=\dfrac{8}{2} \\
 & \Rightarrow x=4 \\
\end{align}\]
Also when we put the denominator of the given equation equal to zero we will get
$\begin{align}
  & \Rightarrow {{\left( x+2 \right)}^{6}}=0 \\
 & \Rightarrow x=-2 \\
\end{align}$

Hence $x=-2$ and $x=4$ are the critical points for the given equation.

Note: We can also find the point of maxima and minima by using the second order derivative. If the value of second order derivative is negative the function has a point of maxima and if the value of second order derivative is positive the function has a point of minima.