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

Question

Isaiah Allen · Accepted Answer

there is only one critical point, at x = 0

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

Christopher Allers · Answer

undefined To find the critical points of $ f(x) = x^{\frac{2}{3}} \cdot e^{x^2 - 4} $, we first need to find its derivative, set it equal to zero, and then solve for $ x $.

$$ f'(x) = \frac{2}{3}x^{-\frac{1}{3}} \cdot e^{x^2 - 4} + 2x \cdot e^{x^2 - 4} $$

Setting $ f'(x) $ equal to zero:

$$ \frac{2}{3}x^{-\frac{1}{3}} \cdot e^{x^2 - 4} + 2x \cdot e^{x^2 - 4} = 0 $$

We can't solve this equation explicitly for $ x $ in terms of elementary functions, but we can find the critical points by setting each factor equal to zero and solving for $ x $.

For the first factor:
$$ \frac{2}{3}x^{-\frac{1}{3}} \cdot e^{x^2 - 4} = 0 $$
$$ x^{-\frac{1}{3}} = 0 $$

This has no real solution because $ x $ cannot be zero.

For the second factor:
$$ 2x \cdot e^{x^2 - 4} = 0 $$
$$ 2x = 0 $$
$$ x = 0 $$

So, the only critical point is $ x = 0 $.