$f(x)=x^4+bx^3+cx^2+dx+e$ where $b,c,d,e$ are real numbers and $3b^2<8c$. Show that the function $f(x)$ has a unique minimum.

Question

QUESTION: Consider the real valued function $f$ defined over the interval $(-\infty,\infty)$ by $f(x)=x^4+bx^3+cx^2+dx+e$ where $b,c,d,e$ are real numbers and $3b^2<8c$. Show that the function $f(x)$ has a unique minimum.

MY APPROACH: $f'(x)=4x^3+3bx^2+2cx+d$. Now at stationary points, $f'(x)=0$, which implies that $4x^3+3b^2+2cx+d=0$. If I can prove that this equation is only satisfied for one and only one real $x$ then the problem is solved. This implies that the discriminant must be negative.. I applied the formula for discriminant of a cubic equation which is $D=a^2b^2+18abc-4b^3-4a^3c-27c^2$. Now after some calculation I arrive at $\frac{9}{16}b^2+\frac{27}4bc-\frac{27}{16}b^3-2c-\frac{27}4c^2$. Now as you can see I am clearly lost here..

Is there a smarter method to think about this one?.. I could not even implement the given condition that $3b^2<8c$. Any help is much appreciated.

Thank you.

Answer 1

For $4x^3+3bx^2+2cx+d=0$ to have only one real solution. The sign of the slope of $f'(x)=4x^3+3bx^2+2cx+d$ should not change, i.e. $f^{''} > 0$ (or $< 0$) for all $x$. Thus we want $12x^2+6bx+2c > 0$ for all $x$. This will happen only when the discriminant $3b^2-8c$ of this quadratic expression is $< 0$.

Answer 2

Suppose there were two (or more) (local) minima. Then there are two (or more) zeroes of the first derivative and, by the mean value theorem (or just Rolle's theorem), a zero of the second derivative, between any two zeroes of the first.

Now $f''(x) = 12 x^2 + 6 b x + 2c$ has discriminant $36b^2 - 96 b = 12(3b^2 - 8 c)$, which should be something you can say something about...