How do you find the inflection points for $ f\left( x \right)={{x}^{4}} $ ?

Hint: We first explain the process of points of inflection. We find the concept of inflection being similar to finding extremum points. The second derivative of the function gives us the solution of the problem.

Complete step-by-step answer:
We have to find the inflection points for $ f\left( x \right)={{x}^{4}} $ . Inflection points are points where the function changes concavity, from being "concave up" to being "concave down" or vice versa. They can be found by considering where the second derivative changes signs.
Therefore, we have to find the second derivative of the function $ f\left( x \right)={{x}^{4}} $ .
We know that \[\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}\]. We use that for the differentiation where $ f\left( x \right)={{x}^{4}} $ .
For our given function $ f\left( x \right)={{x}^{4}} $ , we can differentiate it to $ {{f}^{'}}\left( x \right)=\dfrac{df}{dx} $ .
The value of n is 4. We apply the theorem and get $ \dfrac{df}{dx}=\dfrac{d}{dx}\left( {{x}^{4}} \right)=4{{x}^{3}} $ .
For the second derivative of the function $ {{f}^{'}}\left( x \right)=4{{x}^{3}} $ , we can convert it to \[{{f}^{''}}\left( x \right)=\dfrac{d\left( {{f}^{'}} \right)}{dx}\].
The value of n is 3. We apply the theorem and get $ {{f}^{''}}\left( x \right)=\dfrac{d\left( {{f}^{'}} \right)}{dx}=\dfrac{d}{dx}\left( 4{{x}^{3}} \right)=12{{x}^{2}} $ .
We know that for $ \forall x\in \mathbb{R} $ , the value of $ {{f}^{''}}\left( x \right)=12{{x}^{2}}>0 $ .
Therefore, the function never changes its concavity.

Note: We need to remember the concept of extremum and point of inflection is quite similar. The slopes decide the concavity and its change. We also need to remember that If a function is undefined at some value of x, there can be no inflection point.