planhetkk

2022-09-05

What would the critical points be of this equation?

$\frac{dy}{dx}}={\displaystyle \frac{-4x}{(1+{x}^{2}{)}^{2}}$

I just dont understand how find them with a reciprocal function. If someone could explain how to find it in this this situation that would be fantasic!

$\frac{dy}{dx}}={\displaystyle \frac{-4x}{(1+{x}^{2}{)}^{2}}$

I just dont understand how find them with a reciprocal function. If someone could explain how to find it in this this situation that would be fantasic!

Derick Ortiz

Beginner2022-09-06Added 11 answers

Explanation:

The critical point of a function is a value of x in the domain of f such that either f′(x) is 0 or undefined. From your equation, ${f}^{\prime}(x)=0\to x=0$, and f′(x) is always defined for all x. So $x=0$ is the only critical value.

The critical point of a function is a value of x in the domain of f such that either f′(x) is 0 or undefined. From your equation, ${f}^{\prime}(x)=0\to x=0$, and f′(x) is always defined for all x. So $x=0$ is the only critical value.

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