Answer with explanation:
Drawing the graph of,
[tex]y = x^4 - 2 x^2 + 1\\\\=(x^2-1)^2[/tex]
Used the Identity
(a - b)²=a² - 2 a b + b²
To find the Maximum , we will differentiate the function
y'=2×(x²-1)×(2 x)
y' = 4 x(x²-1)
For maximum or Minimum
y'=0
→4 x(x²-1)=0
→x=0,∧ x²-1=0
→x²=1
→x=1 ∧ x= -1
So, three critical Points are, x=0,1 and -1.
[tex]y(0)=0^4-2\times (0)^2+1\\\\ y(0)=1\\\\y(1)=1^4-2 \times (1)^2+1\\\\y(1)=2-2\\\\y(2)=0\\\\y(-1)=(-1)^4-2 \times (-1)^2+1\\\\y(-1)=1-2+1\\\\y(-1)=0[/tex]
At, x=0,function attains Maximum Value.
→Maximum value of the function ,is 1.
⇒ you can check graphically also.
