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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
(Still high on mods! Next week, will make questions on some other topic.)
(A) 3
(B) 4
(C) 5
(D) 21/4
(E) 7
(Still high on mods! Next week, will make questions on some other topic.)
That's right! I can count on you guys to always give the correct answer and some innovative ways of solving...
Let me give you the method I would use to solve it too...
Where mods are concerned, the fact that mod signifies the distance from 0 on the number line is my best friend...
So |x-3| is the distance from 3, |x + 1| is the distance from -1 and |4x - 1| is 4 times the distance from 1/4
So |4x - 1| + |x-3| + |x + 1| is the sum of distances from 3, -1 and four times the distance from 1/4 e.g. if x takes the value at point A, f(x) will be sum of length of red line, green line and blue line. the question here is, what is the minimum such sum possible?
Ques.jpg [ 4.56 KiB | Viewed 33949 times ]
Can I say that minimum total distance will be covered from point 1/4?
Ques1.jpg [ 3.43 KiB | Viewed 33857 times ]
The logic being that the distance from 3 to -1, which is 4 units has to be covered. Why to cover any distance from 1/4 at all?
Check this video discussing basics of how to use absolute values as distances on the number line: https://youtu.be/oqVfKQBcnrs
Let me give you the method I would use to solve it too...
Where mods are concerned, the fact that mod signifies the distance from 0 on the number line is my best friend...
So |x-3| is the distance from 3, |x + 1| is the distance from -1 and |4x - 1| is 4 times the distance from 1/4
So |4x - 1| + |x-3| + |x + 1| is the sum of distances from 3, -1 and four times the distance from 1/4 e.g. if x takes the value at point A, f(x) will be sum of length of red line, green line and blue line. the question here is, what is the minimum such sum possible?
Attachment:
Ques.jpg [ 4.56 KiB | Viewed 33949 times ]
Can I say that minimum total distance will be covered from point 1/4?
Attachment:
Ques1.jpg [ 3.43 KiB | Viewed 33857 times ]
The logic being that the distance from 3 to -1, which is 4 units has to be covered. Why to cover any distance from 1/4 at all?
Check this video discussing basics of how to use absolute values as distances on the number line: https://youtu.be/oqVfKQBcnrs
30 Oct 2010, 01:59
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Let g(x) = |4x - 1| & h(x)=|x-3| + |x + 1| ... So f(x)=g(x)+h(x)
Now h(x) is equal to 4 between -1 and 3 and is higher everywhere else, so it minimizes between -1 & 3
g(x) is minimum at x=(1/4) where it is equal to 0.
Since -1<(1/4)<3 & f(x)=g(x)+h(x) ... It is straight forward to imply that f(x) will be minimum at 1/4, since both g(x) & h(x) are minimum at that point
f(1/4)=g(1/4)+h(1/4)=0+4
Hence answer is 4
Now h(x) is equal to 4 between -1 and 3 and is higher everywhere else, so it minimizes between -1 & 3
g(x) is minimum at x=(1/4) where it is equal to 0.
Since -1<(1/4)<3 & f(x)=g(x)+h(x) ... It is straight forward to imply that f(x) will be minimum at 1/4, since both g(x) & h(x) are minimum at that point
f(1/4)=g(1/4)+h(1/4)=0+4
Hence answer is 4
