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Find the minimum value of
\frac{x^2}{x + 2}
for x > 2.

 
 Mar 18, 2025
 #1
avatar+130466 
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  x^2  / ( x + 2)  =   x^2 (x + 2)^(-1)  = f(x)

 

Take the derivative and set to 0

 

f'(x)  =  2x (x + 2)^(-1) - x^2 (x + 2)^(-2)  = 0

 

(x + 2)^(-2) [ 2x ( x + 2) - x^2]  =  0

 

x^2 + 4x  =  0

 

x ( x + 4)  = 0

 

Local Max occurs at x = -4

Local Min occurs at x = 0

 

No min for  x >  2

 

cool cool cool

 Mar 19, 2025
edited by CPhill  Mar 19, 2025

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