Question

The minimum value of the function f(x) = 2x4 – 3x2 + 2x – 5, x ∈ [–2, 2] is

Answer 1

The minimum value of the function f(x) = 2x⁴ – 3x² + 2x – 5 is ₋8

Step-by-step explanation:

Given:

Function $f(x)=2x^4-3x^2+2x-5$      $x\in [-2,2]$

To find out:

The minimum value of the function f(x)

Solution:

$f(x)=2x^4-3x^2+2x-5$

$f'(x)=8x^3-6x+2$

Let $p(x)=8x^3-6x+2$

$p(-1)=8(-1)^3-6(-1)+2$

$\implies p(-1)=-8+6+2$

$\implies p(-1)=0$

Therefore, x = -1 is a zero of the polynomial p(x)

$\therefore p(x)=(x+1)(8x^2-8x+2)$

$\implies p(x)=4(x+1)(4x^2-4x+1)$

$\implies p(x)=4(x+1)[(2x)^2-2\times 2x\times 1+1^2]$

$\implies p(x)=4(x+1)(2x-1)^2$

Therefore,

$f'(x)=4(x+1)(2x-1)^2$

For maxima or minima

$f'(x)=0$

$\implies 4(x+1)(2x-1)=0$

This gives us

$x=-1, x=\frac{1}{2}$

Now

$f"(x)=24x^2-6$

At x = -1

$f"(-1)=24-6=18>0$

i.e. at x = -1 the function attains minima in [-2,2]

Therefore, minimum value of the function is at x = -1

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

$\implies f(-1)=2-3-2-5=-8$

Hope this answer is helpful.

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