Question

The absolute minimum value of x4 – x2 – 2x+ 5

(A) is equal to 5 (B) is equal to 3

(C) is equal to 7 (D) does not exist

Answer 1

The abs. minimum value of the given function was found to be 3.

Solution-

Here, the given function is,

`f(x)=x^4-x^2-2x+5`

Then, calculating its first derivative,

`{f}'(x)=4x^3-2x-2`

Then, calculating its second derivative,

`{f}''(x)=12x^2-2`

Then, calculating all the critical values of the given function by equating the first derivative to 0,

`\Rightarrow {f}'(x)=0`

`\Rightarrow 4x^3-2x-2=0`

`\Rightarrow 2x^3-x-1=0`

`\Rightarrow 2x^3-x-1=0`

`\Rightarrow 2x^3-2x+x-1=0`

`\Rightarrow 2x(x^2-1)+(x-1)=0`

`\Rightarrow 2x(x+1)(x-1)+(x-1)=0`

`\Rightarrow (x-1)(2x(x+1)+1)=0`

`\Rightarrow (x-1)(2x^2+2x+1)=0`

`\Rightarrow (x-1)=0` ( ∵ ignoring the imaginary roots)

`\Rightarrow x=1`

Putting the value of x, in the second derivative,

`{f}''(1)=12(1)^2-2=10`

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,

`f(1)=(1)^4-(1)^2-2(1)+5=3`

∴ The abs. minimum value of the given function is 3.