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

1 Answer
Jan 1, 2018

Answer:

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 #2e^(x^2 -4)/(3x^(1/3)) + 2x^(5/3) * 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 ( #2e^(x^2 -4)/(3x^(1/3))# and #2x^(5/3) * e^(x^2 - 4)# )must be equal and opposite in sign

#therefore -2e^(x^2 -4)/(3x^(1/3)) = 2x^(5/3) * e^(x^2 - 4)#

after eliminating the same terms which are #e^(x^2 - 4)# and #2#,

we are left with #-1/(3x^(1/3) ) = x^(5/3)#

multiplying both sides by #x^(1/3)#

#-1/3 = x^2#
# sqrt (-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