Approaching this proof problem? If $0 \le x \le 3$ then $12 - 7x + x^2 \ge 0.$

Question

Prove that if $x$ is a real number in the range $12 - 7x + x^2 \ge 0.$

Which type of proof should I use to solve this? At first I thought direct proof. Choosing a number between $0$ and $3$ and attempting to solve?

Answer 1

Hints:

1. $$12-7x+x^2=\left(x-\frac{7}{2}\right)^2+12-\frac{49}{4}=\left(x-\frac{7}{2}\right)^2-\frac{1}{4}$$

2. $$0\leq x\leq 3\implies \frac{1}{4}\leq \left(x-\frac{7}{2}\right)^2\leq \frac{49}{4}$$

Answer 2

Evaluate the endpoints. What happens when $x = 0$, $x=3$? Can the expression ever be negative? There would have to be a point where is moves from the positive side to the negative side, i.e, where the expression is zero. Compute where that happens, or show that it can't, using the quadratic formula.