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avatar+322 

Find the critical numbers of the function.

f(x)= x^(1/9) - x^(-8/9)

 Apr 1, 2019
 #1
avatar+322 
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I think x=0 might be one of the critical values but I'm not sure

 Apr 1, 2019
 #2
avatar+109725 
+1

I do not know what the definition of a critical value is.

BUT

 

When x=0

 

f(0)= 0^(1/9) - 0^(-8/9)

 

\(=0^{1/9}-\frac{1}{0^{8/9}}\\ =0- undefined\\ =undefined\)

 

So there is probably an asymptote at x=0

 

f(x)= x^(1/9) - x^(-8/9)

 

\(f(x)=x^{1/9}-x^{-8/9}\\ f'(x)=\frac{x^{-8/9}}{9}+\frac{8x^{-17/9}}{9}\\ f'(x)=\frac{1}{9x^{8/9}}+\frac{8}{9x^{17/9}}=0\\ \)

 

By inspection I do not think this is ever the case.

So there are no stationary points.

 

Here is the graph.

 

 Apr 1, 2019

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