y = min ( ∣ x ∣ , 1 + ∣ ∣ x ∣ + x ∣ , 2 + ∣ ∣ x ∣ − x ∣ )
On the interval [ − 2 , 4 ] , what does the graph of y look like?
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Haha. Yea, that's how I came up with the problem.
I wanted to convert it into a single function, but then that became too complicated.
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I like the Heaviside step function for this reason.
y = 1 + ( ∣ x ∣ − 1 ) u ( x + 1 ) + ( 2 − ∣ x ∣ ) u ( x − 2 )
Okay, well maybe it's not any less complicated...
y = min ( ∣ x ∣ , 1 + ∣ ∣ x ∣ + x ∣ , 2 + ∣ ∣ x ∣ − x ∣ ) = ⎩ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎧ x < 0 x ≥ 0 ⇒ min ( ∣ x ∣ , 1 , 2 + 2 ∣ x ∣ ) ⇒ min ( ∣ x ∣ , 1 + 2 ∣ x ∣ , 2 ) = { − 2 ≤ x < − 1 − 1 ≤ x < 0 ⇒ 1 ⇒ ∣ x ∣ = { 0 ≤ x < 2 2 ≤ x ≤ 4 ⇒ ∣ x ∣ ⇒ 2
The final graph is as follows -- a square-root symbol .
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One way to graph this function is to graph all 3 equations. It is useful to note that ∣ ∣ x ∣ + x ∣ = 2 x for x ≥ 0 and 0 for x < 0 , while ∣ ∣ x ∣ − x ∣ = 0 for x ≥ 0 and 2 x for x < 0 .
Then, we can "trace" over the minimum of the three graphs:
Alas, we have a square root symbol!