Question

How do you find the x and y coordinates of all inflection points $f\left( x \right)={{x}^{4}}-12{{x}^{2}}$?

Answer 1

The inflection point of a function is defined as the point where its second derivative is equal to zero, or is undefined. So we need to differentiate the function $f\left( x \right)={{x}^{4}}-12{{x}^{2}}$ two times to obtain the second derivative function, which will be a quadratic polynomial. On equating the second derivative to zero, we will get a quadratic equation. On solving the quadratic equation, we will get the x coordinates of the inflection points. On substituting the x-coordinates in the given function, we will obtain the y-coordinates of inflection points.

Complete step-by-step answer:
The function given in the question is
$f\left( x \right)={{x}^{4}}-12{{x}^{2}}........(i)$
For determining the inflection points of the given function, we have to find the second derivative of the function. On differentiating the above function, we get
$\Rightarrow f'\left( x \right)=4{{x}^{3}}-24x$
Now, we again differentiate the above equation to get the second derivative as
$\begin{aligned} & \Rightarrow f''\left( x \right)=12{{x}^{2}}-24 \\\ & \Rightarrow f''\left( x \right)=12\left( {{x}^{2}}-2 \right) \\\ \end{aligned}$
Writing $2={{\left( \sqrt{2} \right)}^{2}}$ in the above equation, we get
$\Rightarrow f''\left( x \right)=12\left( {{x}^{2}}-{{\left( \sqrt{2} \right)}^{2}} \right)$
Now, we know that ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$. So the above equation can be written as
$\Rightarrow f''\left( x \right)=12\left( x+\sqrt{2} \right)\left( x-\sqrt{2} \right)........(ii)$
Now, the inflection points are the points where the second derivative is zero. That is,
$\Rightarrow f''\left( x \right)=0$
Putting (ii) in the above equation, we get

& \Rightarrow 12\left( x+\sqrt{2} \right)\left( x-\sqrt{2} \right)=0 \\\ & \Rightarrow x+\sqrt{2}=0,x-\sqrt{2}=0 \\\ & \Rightarrow x=-\sqrt{2},x=\sqrt{2} \\\ \end{aligned}$$ Hence, the x-coordinates of the inflection points are $\sqrt{2},-\sqrt{2}$. For the y-coordinates, we substitute $x=\sqrt{2}$ in (i) to get $\begin{aligned} & \Rightarrow f\left( \sqrt{2} \right)={{\left( \sqrt{2} \right)}^{4}}-12{{\left( \sqrt{2} \right)}^{2}} \\\ & \Rightarrow f\left( \sqrt{2} \right)=4-12\left( 2 \right) \\\ & \Rightarrow f\left( \sqrt{2} \right)=-20 \\\ \end{aligned}$ Similarly, we substitute $x=-\sqrt{2}$ in (i) to get $$\begin{aligned} & \Rightarrow f\left( -\sqrt{2} \right)={{\left( -\sqrt{2} \right)}^{4}}-12{{\left( -\sqrt{2} \right)}^{2}} \\\ & \Rightarrow f\left( -\sqrt{2} \right)=4-12\left( 2 \right) \\\ & \Rightarrow f\left( -\sqrt{2} \right)=-20 \\\ \end{aligned}$$ Hence, the y-coordinates of the inflection points are $-20,-20$. Hence, the coordinates of the inflection points are $\left( \sqrt{2},-20 \right)$ and $\left( -\sqrt{2},-20 \right)$. **Note:** Do not confuse the inflection point concept with the extremum point. The extremum points are the points where the first derivative of the function is equal to zero. But the inflexion points are those points where the second derivative is equal to zero or is undefined. Do not equate the first derivative to zero for obtaining the inflection points of the given function.