# What are the critical points of f(x)=x^(2/3) *e^ (x^2-4)?

Jan 1, 2018

there is only one critical point, at x = 0

#### Explanation:

critical points are the points in a function where the slope changes from positive to negative or vice versa, therefore these points have potential to be the maximum or minimum point of a function.

this change of slop can happen at two types of situations, where the derivative is 0 and where it is undefined these places can also be the maximum or minimum value.

to find the critical points, we have to see where the derivative is 0 or undefined so you need to differentiate the function and
the derivative is $2 {e}^{{x}^{2} - 4} / \left(3 {x}^{\frac{1}{3}}\right) + 2 {x}^{\frac{5}{3}} \cdot {e}^{{x}^{2} - 4}$

you can see that it will be undefined at $x = 0$ as x is a denominator in the equation and division by 0 is undefined
so 1 critical point is $x = 0$

to get the derivative equal to zero the two terms ( $2 {e}^{{x}^{2} - 4} / \left(3 {x}^{\frac{1}{3}}\right)$ and $2 {x}^{\frac{5}{3}} \cdot {e}^{{x}^{2} - 4}$ )must be equal and opposite in sign

$\therefore - 2 {e}^{{x}^{2} - 4} / \left(3 {x}^{\frac{1}{3}}\right) = 2 {x}^{\frac{5}{3}} \cdot {e}^{{x}^{2} - 4}$

after eliminating the same terms which are ${e}^{{x}^{2} - 4}$ and $2$,

we are left with $- \frac{1}{3 {x}^{\frac{1}{3}}} = {x}^{\frac{5}{3}}$

multiplying both sides by ${x}^{\frac{1}{3}}$

$- \frac{1}{3} = {x}^{2}$
$\sqrt{- \frac{1}{3}} = x$
now Unless we are Using complex numbers
a square of any number cannot be negative therefore there doesn't exist any real value of x which makes the derivative equal to 0, only undefined at x = 0

therefore the only critical point is where x = 0
you can see this on the graph of the function graph{x^(2/3) * e^(x^2 - 4) [-11.25, 11.25, -5.625, 5.625]}

the only place where the derivative or the slope changes from negative to positive s at x = 0