Respuesta :
the answer is
g'(x)=4x^3-10x=0 so (4x-10)x=0 it is x= 0 and x= 5/2
g(0)=4
g(5/2)=x^4-5x^2+4=11.81
the minimum is M(0, 4)
g'(x)=4x^3-10x=0 so (4x-10)x=0 it is x= 0 and x= 5/2
g(0)=4
g(5/2)=x^4-5x^2+4=11.81
the minimum is M(0, 4)
g ( x ) = x ^4 - 5 x² + 4
g `( x ) = 4 x³ - 10 x
4 x³ - 10 x = 2 x ( 2 x² - 5 ) = 0 ( the local minimum is where the derivative is equal to zero )
x = 0,
2 x² - 5 = 0
x² = 5/2
x = +/- √(5/2) = +/- 1.58
g ( 0 ) = 4 ( not the local minimum )
g ( √5/2 ) = g ( -√5/2 ) = ( √5/2 )^4 - 5 (√5/2 )² + 4 =
= 2.5 ² - 5 · 2.5 + 4 = 6.25 - 12.5 + 4 = - 2.25
Answer:
The local minimum is : M ( - 1.58 , - 2.25 ) and N ( 1.58, - 2.25 ).
g `( x ) = 4 x³ - 10 x
4 x³ - 10 x = 2 x ( 2 x² - 5 ) = 0 ( the local minimum is where the derivative is equal to zero )
x = 0,
2 x² - 5 = 0
x² = 5/2
x = +/- √(5/2) = +/- 1.58
g ( 0 ) = 4 ( not the local minimum )
g ( √5/2 ) = g ( -√5/2 ) = ( √5/2 )^4 - 5 (√5/2 )² + 4 =
= 2.5 ² - 5 · 2.5 + 4 = 6.25 - 12.5 + 4 = - 2.25
Answer:
The local minimum is : M ( - 1.58 , - 2.25 ) and N ( 1.58, - 2.25 ).