Find value of $x$ which minimizes $f(x)=ax^4+bx+c$, where $(a,b,c) \in \mathbb{R}$ and..

Question

Problem

Find value of $x$ which minimizes $f(x)=ax^4+bx+c$, where $(a,b,c) \in \mathbb{R}$ and find conditions a,b and c satisfy in order for the function to have a unique and finite minimum.

Solution

Critical points can be found rather easily when, $f'(x)=0$ which happens when

$$ x=\sqrt[3]{-\frac{b}{4a}} $$

Whether the critical point is minima or maxima it can be determined with sign of second derivative therefore if $f''(x)>0 \implies$ minimum, which means that

$$ 12a\left(\sqrt[3]{-\frac{b}{4a}}\right)^2 > 0 \implies \text{ minimum} $$

In order for the function to have unique and finite minimun conditions, $a \neq 0$ and $a \cdot b \le 0$ must be satisfied. Since $\frac{b}{a} \le 0$ in order to the square root to be real-valued this must be met and dividing with zero is not defined hence $a \neq 0$.

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However the conditions in order to the function have unique and finite minima appears to be wrong? Any explanation how how exactly this is wrong?

Answer 1

\begin{align} &f(x)=ax^4+bx+c\\ \implies &f'(x)=4ax^3+b=0\\ \implies &f'(x_0)=0, x_0=\frac{-b}{4a} \end{align} Further, $f''(x)=12a x^2>0 ~for~ a > 0$. So the mininmum of $f(x)$ is at $x=x_0$ and $f_{min} =f(x_0)=\frac{3b}{4} \left(\frac{-b}{4a}\right)^{1/3}+c,~ a>0.$