Porisma

Worked solution

Mirror the non-negative input side

Use a graphing calculator to sketch (1) y=ln⁡(x+2)y=\ln(x+2) and (2) y=ln⁡(∣x∣+2)y=\ln(|x|+2). Copy the curves. What do you observe about the curves?

CJC · Transformations · Input modulus activity

Sketch and compare the logarithmic curves

Answer

Retain ln⁡(x+2)\ln(x+2) for x≥0x\ge0 and mirror that portion in the y-axis.

The source graph retains the original x-intercept (−1,0)(-1,0), original vertical asymptote x=−2x=-2 and common y-intercept (0,ln⁡2)(0,\ln2). The new graph is symmetric about the y-axis. Independently, ∣x∣+2≥2|x|+2\ge2 means its real domain is all x and its range is [ln⁡2,∞)[\ln2,\infty); the discarded negative-input side does not create an asymptote at x=−2x=-2.

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Original ln(x+2), domain x>−2, asymptote x=−2, root (−1,0), and mirrored ln(|x|+2), defined for all real x with minimum (0,ln2). The new curve has no vertical asymptote.
Original ln(x+2), domain x>−2, asymptote x=−2, root (−1,0), and mirrored ln(|x|+2), defined for all real x with minimum (0,ln2). The new curve has no vertical asymptote.
Original ln(x+2), domain x>−2, asymptote x=−2, root (−1,0), and mirrored ln(|x|+2), defined for all real x with minimum (0,ln2). The new curve has no vertical asymptote.
How did this part go?

Use the piecewise input definition

Answer

f(∣x∣)={f(x),x≥0,f(−x),x<0.f(|x|)=\begin{cases}f(x),&x\ge0,\\f(-x),&x<0.\end{cases}

The source defines ∣x∣=x|x|=x for x≥0x\ge0 and ∣x∣=−x|x|=-x for x<0x<0. Its two steps retain the original portion for x≥0x\ge0 and reflect this retained portion about the y-axis. Thus the result is symmetric about that axis wherever the inputs are valid. The later “Hence to sketch” line omits the modulus bars in both originals; the surrounding definition and steps establish that it refers to f(∣x∣)f(|x|). Keep the original typo in the native transcription.

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