Best solutions to - "Prove that f(f(x))≥0 for all real x"

High School
Algebra
Algebra I
Quadratic function and equation

Destinee Hensley

Answered question

2022-03-29

Prove that

f (f (x)) \geq 0

for all real x

Answer & Explanation

Abdullah Avery

Beginner2022-03-30Added 19 answers

Step 1
As you mention, $a \geq \frac{1}{4}$ works. The only case that remains is when $0 < a < \frac{1}{4}$ , i.e. when f(x) has two distinct real roots. In this case, it is enough to ensure that the midpoint of the roots, i.e. $x = - \frac{1}{2 a}$ , results in a non-negative value, so we check
$f (- t \frac{1}{2 a}) = \frac{1}{4 a} - \frac{1}{2 a} + 1 = 1 - \frac{1}{4 a}$
$\Rightarrow f (f (- t \frac{1}{2 a})) = f (1 - t \frac{1}{4 a}) = \frac{16 a^{2} + 24 a - 3}{16 a}$
and from $16 a^{2} + 24 a - 3 \geq 0$ we get $a \geq d \frac{2 \sqrt{3} - 3}{4}$

Lana Hamilton

Beginner2022-03-31Added 12 answers

Step 1
Firstly,

1 - 4 a \leq 0

is valid and since

a = 0

is not valid, it's enough to check

0 < a < \frac{1}{4}

, which gives

a x^{2} + x + 1 = \frac{- 1 + \sqrt{1 - 4 a}}{2 a}

and

a x^{2} + x + 1 = \frac{- 1 - \sqrt{1 - 4 a}}{2 a}

have no real roots.
It's enough to work with the first equation, which gives:

1 - 4 a (1 - \frac{- 1 + \sqrt{1 - 4 a}}{2 a}) \leq 0

2 \sqrt{1 - 4 a} \leq 4 a + 1,

which gives

a \geq \frac{2 \sqrt{3} - 3}{4} .

It's interesting that in the case

a = \frac{2 \sqrt{3} - 3}{4}

we obtain:

f (f (x)) = \frac{1}{64} {(\sqrt{3} x + 2 \sqrt{3} + 4)}^{2} ((26 \sqrt{3} - 45) x^{2}

+ 4 (7 - 4 \sqrt{3}) x + 44 - 24 \sqrt{3}) \geq 0 .

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