Answer:
given function has 2 minimums - [tex]\frac{9}{4}[/tex] and [tex]\frac{9}{4}[/tex]
Step-by-step explanation:
Step 1. g'(x) = 4x³ - 10x
Step 2. Find find the critical points:
4x³ - 10x = 2x(2x² - 5) = 0
[tex]x_{1}[/tex] = - [tex]\sqrt{\frac{5}{2} }[/tex] , [tex]x_{2}[/tex] = 0 , [tex]x_{3}[/tex] = [tex]\sqrt{\frac{5}{2} }[/tex]
Step 3. g'(x) > 0 : - [tex]\sqrt{\frac{5}{2} }[/tex] < x < 0 or x > [tex]\sqrt{\frac{5}{2} }[/tex]
g'(x) < 0 : x < - [tex]\sqrt{\frac{5}{2} }[/tex] or 0 < x < [tex]\sqrt{\frac{5}{2} }[/tex]
Step 4.
If x ∈ ( - ∞ , - [tex]\sqrt{\frac{5}{2} }[/tex] ) , g(x) is decreasing ;
If x = - [tex]\sqrt{\frac{5}{2} }[/tex] , g(x) has minimum value ;
If x ∈ ( - [tex]\sqrt{\frac{5}{2} }[/tex] , 0 ) , g(x) is increasing ;
If x = 0 , g(x) has maximum value ;
If x ∈ ( 0 , [tex]\sqrt{\frac{5}{2} }[/tex] ) , g(x) is decreasing ;
If x = [tex]\sqrt{\frac{5}{2} }[/tex] , g(x) has minimum value ;
If x ∈ ( [tex]\sqrt{\frac{5}{2} }[/tex] , ∞ ) , g(x) is increasing .
⇒ at ( - [tex]\sqrt{\frac{5}{2} }[/tex] , - [tex]\frac{9}{4}[/tex] ) and at ( [tex]\sqrt{\frac{5}{2} }[/tex] , [tex]\frac{9}{4}[/tex] ) , g(x) reaches its minimum
