# ChallengeDetermine the minimum value

#### anemone

##### Paris la ville de l'amour
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Math Helper
Determine the minimum value of the function $$\displaystyle f(x)=x^4-6x^2+8x-3$$.

MarkFL

#### MarkFL

##### La Villa Strangiato
Staff member
Moderator
Math Helper
I would first examine the roots of the first derivative for turning points:

$$\displaystyle f'(x)=4x^3-12x+8=4\left(x^3-3x+2\right)=0$$

We can see that $$x=1$$ is a root of the cubic factor, and so let's use synthetic division to fully factor:

$$\displaystyle \begin{array}{c|rr}& 1 & 0 & -3 & 2 \\ 1 & & 1 & 1 & -2 \\ \hline & 1 & 1 & -2 & 0 \end{array}$$

And so:

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

And so our critical values are:

$$\displaystyle x\in\{-2,1\}$$

Because the root of $$f'(x)$$ defined by $$x=1$$ is of odd multiplicity, this is not a turning point, since the sign of the derivative does not change across it, and so it cannot be a local extremum. And so to determine the nature of the sole extremum, we may observe that:

$$\displaystyle f''(x)=4\left(3x^2-3\right)=12(x+1)(x-1)$$

And then:

$$\displaystyle f''(-2)>0$$

Thus, by the second derivative test, the extremum is a minimum. And since:

$$\displaystyle \lim_{x\to\pm\infty}f(x)=\infty$$

We know this minimum is a global minimum, hence:

$$\displaystyle f_{\min}=f(-2)=-27$$

anemone

Staff member