Porisma

Worked solution

Mirror a supplied two-branch sketch

Sketch y=f(∣x∣)y=f(|x|) given the graph of y=f(x)y=f(x).

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Given schematic graph y=f(x): two branches, vertical asymptote x=1, horizontal asymptote y=1, y-intercept (0,3) and x-intercept (3,0). Left branch increases to positive infinity at 1; right branch increases from negative infinity toward y=1. No formula is supplied.
Given schematic graph y=f(x): two branches, vertical asymptote x=1, horizontal asymptote y=1, y-intercept (0,3) and x-intercept (3,0). Left branch increases to positive infinity at 1; right branch increases from negative infinity toward y=1. No formula is supplied.
Given schematic graph y=f(x): two branches, vertical asymptote x=1, horizontal asymptote y=1, y-intercept (0,3) and x-intercept (3,0). Left branch increases to positive infinity at 1; right branch increases from negative infinity toward y=1. No formula is supplied.

CJC · Transformations · Example 16

Retain the given positive-input features

Answer

The new sketch has vertical asymptotes x=±1x=\pm1, horizontal asymptote y=1y=1, and roots x=±3x=\pm3.

The given sketch has vertical asymptote x=1x=1, horizontal asymptote y=1y=1, y-intercept 3 and x-intercept 3. Retain only its x≥0x\ge0 portion, then mirror it in the y-axis. The Teacher fills the blank Student answer axes. Both sides near each new vertical asymptote must retain the source branch directions.

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Symmetric reconstruction of f(|x|) from the given graph: vertical asymptotes x=−1 and 1, horizontal asymptote y=1, roots −3 and 3, central point (0,3), positive central branches tending upward to the poles, outer branches tending downwards at poles and approaching y=1 from below.
Symmetric reconstruction of f(|x|) from the given graph: vertical asymptotes x=−1 and 1, horizontal asymptote y=1, roots −3 and 3, central point (0,3), positive central branches tending upward to the poles, outer branches tending downwards at poles and approaching y=1 from below.
Symmetric reconstruction of f(|x|) from the given graph: vertical asymptotes x=−1 and 1, horizontal asymptote y=1, roots −3 and 3, central point (0,3), positive central branches tending upward to the poles, outer branches tending downwards at poles and approaching y=1 from below.
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Distinguish a corner minimum from a stationary point

Answer

(0,3)(0,3) is a corner local minimum, not a differentiable stationary point.

The Teacher note says “(0,3) is not a turning point in this example.” Preserve that note with a terminology qualification: the supplied retained branch increases immediately to the right of 0, and its mirror decreases immediately to the left. Consequently the centre is a local minimum in the usual local-extremum sense; the corner is not a differentiable stationary point. Reading the note as denying any local minimum would contradict the drawn graph. No unique formula for f is given or inferred.

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