Home > Calculus > Identifying Stationary Points (Critical Points) for a Function

What are the critical points of #g(x)=x/3 + x^-2/3#?

Answer 1

Aurora Alvarado

#x_c=root(3)2# is a local minimum

We can find the critical points by equating the first derivative of #g(x)# to zero:

#g'(x) =1/3 -2/3x^(-3) = 0#

#1/3 =2/(3x^3)#

#x^3= 2#

#x_c=root(3)2#

We can easily see that #g'(x) < 0# for #x< x_c# and viceversa #g'(x) > 0# for #x>x_c# , so #x_c is a local minimum.

graph{x/3+x^(-2)/3 [-10, 10, -5, 5]}

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Elijah Abram

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

The derivative of ( g(x) ) with respect to ( x ) is given by:

[ g'(x) = \frac{d}{dx}\left(\frac{x}{3} + x^{-\frac{2}{3}}\right) ]

[ = \frac{1}{3} - \frac{2}{3}x^{-\frac{5}{3}} ]

To find critical points, set ( g'(x) = 0 ):

[ \frac{1}{3} - \frac{2}{3}x^{-\frac{5}{3}} = 0 ]

[ \frac{1}{3} = \frac{2}{3}x^{-\frac{5}{3}} ]

[ x^{-\frac{5}{3}} = \frac{1}{2} ]

[ x^{\frac{5}{3}} = 2 ]

[ x = 2^{\frac{3}{5}} ]

So, the critical point of ( g(x) ) is ( x = 2^{\frac{3}{5}} ).

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.