Q. If f(x) = |4x - 1| + |x-3| + |x + 1|, what is the minimum value of f(x)?
(A) 3
(B) 4
(C) 5
(D) 21/4
(E) 7
Ans-B
(A) 3
(B) 4
(C) 5
(D) 21/4
(E) 7
Ans-B
|a-b| = the distance between a and b.[email protected] wrote:Q. If f(x) = |4x - 1| + |x-3| + |x + 1|, what is the minimum value of f(x)?
(A) 3
(B) 4
(C) 5
(D) 21/4
(E) 7
Ans-B
|a+b| = |a-(-b)| = the distance between a and -b.
f(x) = |4x - 1| + |x-3| + |x + 1| = the SUM of 3 distances on the number line.
To MINIMIZE the value of f(x), we must MINIMIZE the 3 distances.
|x+1| + |x-3|:
|x-3| = the distance between x and 3:
|x+1| = the distance between x and -1.
To minimize the sum of these 2 distances, x must be AS CLOSE AS POSSIBLE to -1 and 3.
In other words, x must be BETWEEN -1 and 3, inclusive.
-1<--|x+1|-->x<--|x-3|-->3
Here, since the total distance between the two endpoints is 4, the sum of the 2 distances in red is 4.
To illustrate:
If x=-1, then |x+1| + |x-3| = 4.
If x=0, then |x+1| + |x-3| = 4.
If x=3, then |x+1| + |x-3| = 4.
In each case, since x is between -1 and 3, inclusive, the sum of the two distances is 4:
|x+1| + |x-3| = 4.
If x is positioned OUTSIDE -1 and 3, the sum of the two distances will INCREASE.
If x=-2, then |x+1| + |x-3| = 6.
If x=4, then |x+1| + |x-3| = 6.
Thus:
The LEAST possible value of |x+1| + |x-3| is yielded when x is between -1 and 3, inclusive:
|x+1| + |x-3| = 4.
|4x-1|:
Since |4x-1| cannot be negative, the least possible value of this term is 0.
4x-1 = 0
x=1/4.
Since x=1/4 is between -1 and 3, inclusive, it will also minimize the value of |x+1| + |x-3|.
Thus, when x=1/4, the least possible value for f(x) is yielded:
f(x) = |4x-1| + |x-3| + |x+1| = 0 + 11/4 + 5/4 = 4.
The correct answer is B.
