Question

What are the critical numbers of $f\left( x \right)=1-\dfrac{x}{3-{{x}^{2}}}$?

Answer 1

We are given with the function $f\left( x \right)=1-\dfrac{x}{3-{{x}^{2}}}$ to find out the critical values. Critical values are nothing but the values at which the graph curve is ${{f}^{\grave{\ }}}\left( x \right)=0$. So basically, we have find the derived value of $f\left( x \right)=1-\dfrac{x}{3-{{x}^{2}}}$and find a suitable value that gives us ${{f}^{\grave{\ }}}\left( x \right)=0$.

Complete step-by-step answer:
Let us have brief information regarding the critical values now.
Critical point: A critical point of a function of a single variable, $f\left( x \right)$, is a value ${{x}_{0}}$ in the domain of $f$ where it is not differentiable or its derivative is $0$. A critical value is the image under $f$ of a critical point.
Now let us start finding out the critical numbers of the given function $f\left( x \right)=1-\dfrac{x}{3-{{x}^{2}}}$.
As we know that, critical numbers are obtained only when ${{f}^{\grave{\ }}}\left( x \right)=0$.
So let us find out the ${{f}^{\grave{\ }}}\left( x \right)$ of the given $f\left( x \right)$.
The given $f\left( x \right)$ is in the form of $\dfrac{u}{v}$.
The general rule for $\dfrac{u}{v}$ is $\dfrac{u}{v}=\dfrac{v{{u}^{\grave{\ }}}-u{{v}^{\grave{\ }}}}{{{v}^{2}}}$
Now let us solve our given function by applying the general formula.
Consider $u$ as $x$ and $v$ as $3-{{x}^{2}}$.
On deriving, we get $-\dfrac{3+{{x}^{2}}}{{{\left( 3-{{x}^{2}} \right)}^{2}}}$
Since we are supposed to equate the ${{f}^{\grave{\ }}}\left( x \right)$to$0$, let us equate it and obtain the answer.

& -\dfrac{3+{{x}^{2}}}{{{\left( 3-{{x}^{2}} \right)}^{2}}}=0 \\\ & \Rightarrow -3-{{x}^{2}}=0 \\\ & \Rightarrow 3+{{x}^{2}}=0 \\\ & \Rightarrow {{x}^{2}}=-3 \\\ & \therefore x=\pm \sqrt{3} \\\ \end{aligned}$$ Upon finding it, we obtain $${{f}^{\grave{\ }}}\left( x \right)=\pm \sqrt{3}$$ Since we do not have real values, this function does not have critical points in real domain. **Note:** We must note that the function goes up to infinity at $$x=\pm \sqrt{3}$$ when the denominator is zero. All local extreme are also critical points, but not all of them. When there exist no critical points, it means that there is no change in the slope from positive to negative or vice versa. On plotting the graph for the function, we get