Question

How do you find the local extremas for $g(x)={{x}^{2}}+1$ ?

Answer 1

Hint: To find the local extrema of the given function first we will differentiate the given function with respect to x to find its first order derivative. Then we will equate the first order derivative with zero to find the point of local extrema.

Complete step-by-step answer:
We have been given a function $g(x)={{x}^{2}}+1$.
We have to find the local extremas for the given function.
We know that local extrema of a function is the point at which a function has maximum or minimum value.
Now, first we will differentiate the given function with respect to x to find the first order derivative of the function. Then we will get
\[\begin{align}
  & \Rightarrow g(x)=\dfrac{d}{dx}\left( {{x}^{2}}+1 \right) \\
 & \Rightarrow g(x)=\dfrac{d}{dx}{{x}^{2}}+\dfrac{d}{dx}1 \\
\end{align}\]
Now, we know that $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$
Now, applying the differentiation rule to the above obtained equation we will get
\[\begin{align}
  & \Rightarrow g'(x)=2x+0 \\
 & \Rightarrow g'(x)=2x \\
\end{align}\]
Now, equating the first order derivative with zero we will get
$\begin{align}
  & \Rightarrow g'(x)=0 \\
 & \Rightarrow 2x=0 \\
\end{align}$
Now, simplifying the above obtained equation we will get
$\Rightarrow x=0$
It means the function has local extrema i.e. either minima or maxima at $x=0$ .
Hence $x=0$ is the local extrema of a given function.

Note: The point to be noted is that the first order derivative of a function is the slope of the function. When the slope of the function will be zero the function obtains a maximum or minimum value at that point. In order to find whether it is maxima or minima we need to find the second order derivative of the function.