Answer-
The abs. minimum value of the given function was found to be 3.
Solution-
Here, the given function is,
[tex]f(x)=x^4-x^2-2x+5[/tex]
Then, calculating its first derivative,
[tex]{f}'(x)=4x^3-2x-2[/tex]
Then, calculating its second derivative,
[tex]{f}''(x)=12x^2-2[/tex]
Then, calculating all the critical values of the given function by equating the first derivative to 0,
[tex]\Rightarrow {f}'(x)=0[/tex]
[tex]\Rightarrow 4x^3-2x-2=0[/tex]
[tex]\Rightarrow 2x^3-x-1=0[/tex]
[tex]\Rightarrow 2x^3-x-1=0[/tex]
[tex]\Rightarrow 2x^3-2x+x-1=0[/tex]
[tex]\Rightarrow 2x(x^2-1)+(x-1)=0[/tex]
[tex]\Rightarrow 2x(x+1)(x-1)+(x-1)=0[/tex]
[tex]\Rightarrow (x-1)(2x(x+1)+1)=0[/tex]
[tex]\Rightarrow (x-1)(2x^2+2x+1)=0[/tex]
[tex]\Rightarrow (x-1)=0[/tex] ( ∵ ignoring the imaginary roots)
[tex]\Rightarrow x=1[/tex]
Putting the value of x, in the second derivative,
[tex]{f}''(1)=12(1)^2-2=10[/tex]
As, the value of f"(x) is positive, the function attains its minimum value at x=1.
So, f(1) will give the absolute minimum value of the function,
[tex]f(1)=(1)^4-(1)^2-2(1)+5=3[/tex]
∴ The abs. minimum value of the given function is 3.
