Buktikan bahwa fungsi berikut merupakan involusi:
- g(x) = (k – x3)1/3.
- f(x) = –ln[tanh(x/2)].
Catatan. Di bawah ini adalah grafik fungsi f(x) = –ln[tanh(x/2)].
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Bandung, 19-05-2017
Masih tentang Involusi
Blog Matematika ala Hendra Gunawan
Bermatematika 
Kita harus tunjukkan bahwa f(f(x))=x
Perhatikan:
f(f(x))= f((k – x^3)^1/3) = (k – (k – x^3)^1/3)^3)^1/3 = (x^1/3)^3 = x
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1. Misal y = g(x) = (k – x^3)^1/3
Maka y^3 = k – x^3
x^3 = k – y^3
x = (k – y^3)^1/3
shg g-1(x) = (k – x^3)^1/3 = g(x)
Jadi g(x) adl fungsi involusi.
2. tanh x/2 = (e^x – 1)/(e^x + 1)
Misal y = f(x) = -ln((e^x – 1)/(e^x + 1))
Maka – y = ln((e^x – 1)/(e^x + 1))
(e^x – 1)/(e^x + 1) = e^-y
e^x = (e^-y + 1)/(- e^-y + 1)
e^-x = (- e^-y + 1)/(e^-y + 1)
e^-x = (e^-y – 1)/(e^y + 1)
– x = ln((e^-y – 1)/(e^y + 1))
shg f-1(x) = -ln((e^-y – 1)/(e^y + 1)) = -ln(tanh x/2) = f(x)
Jadi f(x) adl fungsi involusi.
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Mantap, Daffa!
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