What are the values and types of the critical points, if any, of #f(x) =x^2-sqrtx#?

Question

Scarlett Allison · Accepted Answer

#x=2root(3)(2)# is the minimum.

#f(x) =x²-sqrtx# We search #f'(x)=0# #f'(x)=2x+1/(2sqrt(x))# #2x+1/(2sqrt(x))=0# #2xsqrtx=-1/2# #xsqrtx=-1/4# #x³=1/16# #x=2root(3)(2)# Let #X=sqrtx#, we can see #f(X)=X⁴-X#, and so that #x=2root(3)(2)# as a minimum of #f(x)# (because f take the sign of his monoma of higher degree.) \0/ here's our answer!

Samantha Adkison · Answer

undefined To find the critical points of $ f(x) = x^2 - \sqrt{x} $, we first need to find its derivative and then solve for where the derivative equals zero.

$ f'(x) = 2x - \frac{1}{2\sqrt{x}} $

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

$ 2x - \frac{1}{2\sqrt{x}} = 0 $

$ 2x = \frac{1}{2\sqrt{x}} $

$ 4x^2 = 1 $

$ x^2 = \frac{1}{4} $

$ x = \pm\frac{1}{2} $

The critical points are $ x = -\frac{1}{2} $ and $ x = \frac{1}{2} $.

Now, to determine the types of these critical points, we can use the second derivative test or examine the behavior of the function around these points.

Taking the second derivative of $ f(x) $:

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

Evaluating the second derivative at $ x = -\frac{1}{2} $:

$ f''\left(-\frac{1}{2}\right) = 2 + \frac{1}{4\left(-\frac{1}{2}\right)^{3/2}} $

$ f''\left(-\frac{1}{2}\right) = 2 - \frac{1}{\sqrt{-1/2}} $

Since the square root of a negative number is imaginary, $ f''\left(-\frac{1}{2}\right) $ is undefined, indicating that the second derivative test fails at this critical point.

Evaluating the second derivative at $ x = \frac{1}{2} $:

$ f''\left(\frac{1}{2}\right) = 2 + \frac{1}{4\left(\frac{1}{2}\right)^{3/2}} $

$ f''\left(\frac{1}{2}\right) = 2 + \frac{1}{1/2} $

$ f''\left(\frac{1}{2}\right) = 2 + 2 $

$ f''\left(\frac{1}{2}\right) = 4 $

Since the second derivative is positive at $ x = \frac{1}{2} $, it indicates that the function has a local minimum at this point.

Therefore, the values and types of the critical points of $ f(x) = x^2 - \sqrt{x} $ are:

1. Critical point at $ x = -\frac{1}{2} $ is undefined.
2. Critical point at $ x = \frac{1}{2} $ is a local minimum.