Find the maximum for x>0.

01 / Standard#Your turn

f(x)=\frac{3x}{x+3},\quad g(x)=\frac{f(x^2)}x

Official paper · jm01-2026 · II.5(b) · PDF 5

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Substitute x² before dividing by x.

Hint 2

Apply AM–GM to x+3/x.

Worked solution

Simplify using x>0.
$g(x)=\frac{3x}{x^2+3}=\frac3{x+3/x}$
Minimise the positive denominator and identify equality.
$x+\frac3x\ge2\sqrt3,\quad g(x)\le\frac{\sqrt3}2,\quad x=\frac3x\iff x=\sqrt3$

Maximum √3/2 at x=√3.

Checks and common pitfalls: The equality condition is needed to show the upper bound is attainable.

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
Simplify using x>0.
$g(x)=\frac{3x}{x^2+3}=\frac3{x+3/x}$
Minimise the positive denominator and identify equality.
$x+\frac3x\ge2\sqrt3,\quad g(x)\le\frac{\sqrt3}2,\quad x=\frac3x\iff x=\sqrt3$

Think first. Reveal a hint when the class is ready.

2026 JM01