Find the intervals of increase and decrease of the function $f(x)=ax^2+bx+c$.
The problem is from the book "Problems in Calculus of One Variable" by I.A Maron. The solution given in the book is as follows:
Isolating the perfect square from the square trinomial , we have $f(x)=a(x+\frac{b}{2a})^2+\frac {4ac-b^2}{4a}$. If $a>0$ then the function $f(x)$ will increase at those values of $x$ satisfying the inequality $x+\frac {b}{2a}>0$ i.e $x>-\frac {b}{2a}$ and decrease when $x+\frac {b}{2a}<0$ i.e $x<-\frac {b}{2a}$. Thus if $a>0$ the function $f(x)$ decreases in the interval $(-\infty, -\frac {b}{2a})$ and increases in the interval $(-\frac {b}{2a},\infty)$; at $x=-\frac {b}{2a}$ the function $f(x)$ takes the minimum value $\frac {4ac-b^2}{4a}$.Similarly, if $a<0$ the function $f(x)$ increases in the interval $(-\infty, -\frac {b}{2a})$ and decreases in the interval $(-\frac {b}{2a},\infty)$; at $x=-\frac {b}{2a}$ the function $f(x)$ takes the maximum value $\frac {4ac-b^2}{4a}$.
However, I am not getting how are they saying in case $a>0$ that $f(x)$ decreases in the interval $(-\infty, -\frac {b}{2a})$ and increases in the interval $(-\frac {b}{2a},\infty)$ . Same goes for the case, $a<0$. I mean how are they concluding it ? I need an explanation of this without using calculus …I am not quite getting it…Maybe there are similar threads in this site concerning the same topic but I can't seem to find it either…
Best Answer
You can observe that, if $a>0$ then we have:
$$\begin{align}f(x)=\underbrace{a\left(x+\frac{b}{2a}\right)^2}_{\geq 0}+\underbrace{\frac {4ac-b^2}{4a}}_{\text {constant term}}&\ge \frac {4ac-b^2}{4a}\end{align}$$
This implies that, if $x_2>x_1\geq-\frac {b}{2a}$, then: $$\begin{align}&x_2+\frac {b}{2a}>x_1+\frac {b}{2a}\ge 0\\ \implies &\left(x_2+\frac {b}{2a}\right)^2>\left(x_1+\frac {b}{2a}\right)^2\\ \implies &f(x_2)>f(x_1)\end{align}$$
Similarly, if $a<0$ we have:
$$\begin{align}f(x)=\underbrace{a\left(x+\frac{b}{2a}\right)^2}_{\leq 0}+\underbrace{\frac {4ac-b^2}{4a}}_{\text {constant term}}&\leq \frac {4ac-b^2}{4a}\end{align}$$
and note that if $-\frac {b}{2a}\leq x_2<x_1$, then you get:
$$\begin{align}&0\leq x_2+\frac {b}{2a}<x_1+\frac {b}{2a}\\ \implies &\left(x_2+\frac {b}{2a}\right)^2< \left(x_1+\frac {b}{2a}\right)^2\\ \implies &f(x_2)<f(x_1).\end{align}$$