What is the local minimum value of the function g (x) = x^4-5x^2+4? (Round answer to the nearest hundredth)

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) = 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).