Graph this function and identify the local minimum and maximum:
f(x)=2x^2-x^4

Question

Graph this function and identify the local minimum and maximum:
f(x)=2x^2-x^4

anonymous · Answer

get the first derivative equate to 0

campbell_st · Answer

you will need to find the 1st and 2nd derivatives 

$$f(x) = 2x^2 - x^4$$

$$f'(x) = 4x - 4x^3$$

$$f"(x) = 4 - 12x^2$$

solve f'(x) = 0 this will give the stationary points

$$4x - 4x^3 = 0$$

$$4x(1 -x)(1+x) = 0$$

x = -1, 0, 1

these the stationary points in the 2nd derivative to determine maximum, minimum or points of inflexion

f"(x) > 0  minimum

f"(x) < 0 maximum

f"(0) = 0 point of inflexion

anonymous · Answer

"these the stationary points in the 2nd derivative to determine maximum, minimum or points of inflexion 
f"(x) > 0 minimum 
f"(x) < 0 maximum 
f"(0) = 0 point of inflexion"
I don't understand this, can you explain it again?

campbell_st · Answer

well the stationary points from the 1st derivative are

x = -1, 0, 1

the find f"(-1) = 4 - 12(-1)^2 
                         = - 8    
then a maximum exists at (-1, 1)  since f"(-1) < 0

next f"(0) = 4 - 12(0)^2    
                 = 4

therefore a minumum at (0, 0) since f"(0) > 0

and f"(1) = 4 - 12(1)^2
               = -8  

therefore a maximum since f"(x) > 0

its the best way to check stationary points.... 

the alternative method is to pick points close to the stationary points and check for a change in the gradients

-ve, 0 , +ve is a minimum and +ve, 0, -ve is a maximum
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the points of inflexion occur ( chnage in concavity) where

f"(x) = 0

or 0 = 4 - 12x^2

$$  x = \pm \frac{1}{\sqrt{3}}$$