What are the critical points of # f(x)=x^(2/3)x^2 - 16#?

Question

Emma Aldridge · Accepted Answer

#c=0#

Critical points (#c#) occur when #f'(c)=0# or when #f'(c)# doesn't exist.

In order to find #f'(x)#, first simplify #f(x)#.

#f(x)=x^(2/3)x^2-16#

#f(x)=x^(2/3)x^(6/3)-16#

#f(x)=x^(8/3)-16#

Use the power rule: #d/dx[x^n]=nx^(n-1)#

#f'(x)=8/3x^(5/3)#

#f'(x)=0# when #x=0# and is never undefined.

Thus, #c=0#.

graph{x^(2/3)x^2-16 [-32.06, 32.9, -20.27, 12.2]}

Paisley Johnson · Answer

undefined To find the critical points of $ f(x) = x^{\frac{2}{3}}x^2 - 16 $, we first need to find the derivative of $ f(x) $ and then solve for $ x $ where the derivative is equal to zero or undefined.

The derivative of $ f(x) $ is $ f'(x) = \frac{2}{3}x^{-\frac{1}{3}} \cdot 2x + 2x^2 $.

Simplifying, we get $ f'(x) = \frac{4}{3}x^{\frac{2}{3}} + 2x^2 $.

Setting $ f'(x) $ equal to zero to find the critical points:

$$ \frac{4}{3}x^{\frac{2}{3}} + 2x^2 = 0 $$

$$ \frac{4}{3}x^{\frac{2}{3}} = -2x^2 $$

$$ x^{\frac{2}{3}} = -\frac{3}{2}x^2 $$

$$ x = -\frac{3}{2}x^{\frac{8}{3}} $$

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

$$ -\frac{2}{3} = x^{\frac{5}{3}} $$

$$ (-\frac{2}{3})^{\frac{3}{5}} = x $$

$$ x = -\frac{2^3}{3^3} $$

$$ x = -\frac{8}{27} $$

So, the critical point of $ f(x) $ is $ x = -\frac{8}{27} $.