Absolute Value
Lesson · Beginner
Algebra: Functions
For a real number x,
Geometrically, |x| is the distance from x to 0 on the number line.
More generally,
is the distance between x and a.
For real numbers x, y,
This is the Triangle Inequality for real numbers.
A useful equivalent form is
Case 1
If a < 0, there is no solution.
If a ≥ 0,
Case 2
with a ≥ 0
Then
or
Solve
Either
or
Thus
or
Then
For a > 0,
means
while
means
Similarly,
means
Solve
Solution
Add 5:
Divide by 2:
The graph consists of
and
It has a vertex at
and is symmetric about the y-axis.
For
consider
Split into four quadrants.
In the first quadrant,
In the second,
In the third,
In the fourth,
So the graph is a square with vertices
For your example,
the vertices are
The region
is the interior of that square.
Its diagonals both have length
Therefore,
so
For
the area is
For complicated absolute value expressions, split the number line at the points where the inside expressions change sign.
For example,
changes form at
So consider:
Solve
For x ≥ 2,
so
and
For
we have
so there are no solutions.
For x ≤ -1,
Thus
so
Therefore,
Expressions such as
are minimized at a median of the numbers
This is a powerful advanced extension.
For example,
is minimized at
WLOG assume
If n is odd, n = 2k + 1
Equality holds when
If n is even, n = 2k. Similarly
and equality holds when
and
Find the minimum value of
for real x.
The expression is the sum of the distances from x to
For three points on a line, the sum of distances is minimized at the median.
Thus
and the minimum value is
Find the area of the region
This is a diamond centered at
with vertices
The two diagonals each have length
Therefore,
Find the maximum value of
for
For 1 ≤ x ≤ 5,
Outside that interval, the expression increases as x moves farther away.
At x = 0,
At x = 7,
Therefore, the maximum value is
Solve
The critical points are
Case 1: x ≤ -2
and
so
valid.
Case 2: -2 ≤ x ≤ 3/2
not valid.
Case 3: x ≥ 3/2
valid.
Therefore,
Find the minimum value of
There are 32 numbers
So,