Question

Find the local maxima of the function \[f\left( x \right) = \dfrac{6}{{{x^2} + 2}}\]

Answer 1

Hint:
Here, we are required to find the local maxima of the given function \[f\left( x \right)\]. We will find the derivative of the given function and then equate it to zero to find the value of \[x\]. Substituting the value of \[x\] in the given function \[f\left( x \right)\] will give us the required local maxima of the given function.

Complete Step by Step Solution:
In order to find the local maxima of the given function: \[f\left( x \right) = \dfrac{6}{{{x^2} + 2}}\],
First of all, we will find its derivative.
We know that if \[f\left( x \right) = {x^n}\] then its derivative is \[f'\left( x \right) = n{x^{n - 1}}\].
Therefore, if \[f\left( x \right) = \dfrac{1}{x} = {x^{ - 1}}\] then,
 \[\begin{array}{l}f'\left( x \right) = - 1{x^{ - 1 - 1}}\\ \Rightarrow f'\left( x \right) = - {x^{ - 2}} = \dfrac{{ - 1}}{{{x^2}}}\end{array}\]
Hence, differentiating both sides of \[f\left( x \right) = \dfrac{6}{{{x^2} + 2}}\] with respect to \[x\], we get,
\[f'\left( x \right) = \dfrac{{ - 6}}{{{{\left( {{x^2} + 2} \right)}^2}}} \times 2x\]
Multiplying the terms, we get
\[ \Rightarrow f'\left( x \right) = \dfrac{{ - 12x}}{{{{\left( {{x^2} + 2} \right)}^2}}}\]
Now, the next step involved equating this derivative of the given function to 0
\[\begin{array}{l} \Rightarrow f'\left( x \right) = 0\\ \Rightarrow \dfrac{{ - 12x}}{{{{\left( {{x^2} + 2} \right)}^2}}} = 0\end{array}\]
On cross multiplication, we get
\[ \Rightarrow - 12x = 0\]
Dividing both sides by \[ - 12\], we get
\[ \Rightarrow x = 0\]
Therefore, at \[x = 0\] we will get the local maxima.
Hence, substituting \[x = 0\] in the given function \[f\left( x \right)\], we get,
\[f\left( x \right) = \dfrac{6}{{{x^2} + 2}} = \dfrac{6}{{0 + 2}}\]
Adding the terms in the denominator, we get
\[ \Rightarrow f\left( x \right) = \dfrac{6}{2}\]
Dividing the terms, we get
\[ \Rightarrow f\left( x \right) = 3\]
Therefore, the local maxima of the given function is at \[f\left( x \right) = 3\]

Hence, this is the required answer.

Note:
In mathematics, the maxima and minima of a function are the largest and the smallest possible values of a given function. They are collectively known as extrema and they either give these largest and smallest values within a given range or on the entire domain. Maxima and minima are called ‘local’ when it is the largest and smallest value of the function respectively in the given range instead of the entire domain. Whereas, it is called ‘absolute’ when it is on the entire domain.