Solve f(x)≥0 when a=8.

01 / Standard#Your turn

f(x)=a-|x-3|-|x-7|

Official paper · jm01-2024 · II.4(a) · PDF 4

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

The expression is a sum of distances from 3 and 7.

Hint 2

Split at 3 and 7 or use symmetry about 5.

Worked solution

Between 3 and 7 the distance sum is 4; outside it is 2|x−5|.
$|x-3|+|x-7|=\max(4,2|x-5|)$
Require that sum not exceed 8.
$|x-5|\le4\iff1\le x\le9$

x∈[1,9].

Checks and common pitfalls: Include endpoints because the inequality is non-strict.

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
Between 3 and 7 the distance sum is 4; outside it is 2|x−5|.
$|x-3|+|x-7|=\max(4,2|x-5|)$
Require that sum not exceed 8.
$|x-5|\le4\iff1\le x\le9$

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

2024 JM01