Find the range of $f(x) = 3x^4 - 16x^3 + 18x^2 + 5$ without applying differential calculus

Question

Find the range of $f(x) = 3x^4 - 16x^3 + 18x^2 + 5$ without applying differential calculus.

I tried to express $$f(x)=3x^4-16x^3+18x^2+5=A(ax^2+bx+c)^2+B(ax^2+bx+c)+C $$ which is a quadratic in $ax^2+bx+c$ which itself is quadratic in $x$. Comparing coefficients, we get

$$Aa^2=3 \tag{1}$$

$$2abA=-16$$

$$A(b^2+2ac)+aB=18$$

$$2bcA+bB=0$$

$$Ac^2+Bc+C=5$$

But I felt its very lengthy to solve these equations. Any hints?

Answer 1

The range is $[k,+\infty)$ where $k$ is the minimum value such that the inequality $$ 3x^4-16x^3+18x^2+5\ge k $$ is true for any $x \in \mathbb{R}$ and this is the minimum value $k$ such that the equation $$ 3x^4-16x^3+18x^2+5- k=0 $$ has a double solution, that is the discriminant of $3x^4-16x^3+18x^2+5- k$ is null.

The calculation of the discriminant for a quartic polynomial is a bit ''heavy'', but Wolfram Alpha gives:

$$\Delta= -6912(k-5)(k-10)(k+22)$$

so the minimum value is $k=-22$.

Answer 2

Let us consider the minimum of $f(x)$ as $k$. Then $f(x)-k$ has at least one repeated root. That is

$$f(x)-k=3x^4-16x^3+18x^2+5- k=3(x-a)^2(x^2+bx+c)$$

Expand and equate coefficients, you get $k=5-3a^2 c$, and

$$3(b-2a)=-16 \\ 3(c+a^2-2ab)=18 \\ a(ab-2c)=0$$

From third equation if $a=0$, then $k=5$. For $a \neq 0$ combinig three equations gives $a^2=4a-3$. Thus $a=1, 3$.

$a=1$ leads to $c=-\frac{5}{3}$, and $k=5-3(-\frac{5}{3})=10$.

$a=3$ gives $c=1$, and $k=5-27=-22$.

Therefore minimum is $k=-22$.