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Matematica - Ecuatii si inecuatii. Functii

Ecuatii si inecuatii. Functii - Intrebarea 11

Să se determine inversa funcției bijective f colon open parentheses 0 comma infinity close parentheses rightwards arrow open parentheses 2 comma infinity close parentheses comma space f left parenthesis x right parenthesis equals x squared plus 2.

f to the power of negative 1 end exponent colon open parentheses 2 comma infinity close parentheses rightwards arrow open parentheses 0 comma infinity close parentheses comma space f to the power of negative 1 end exponent left parenthesis x right parenthesis equals square root of x minus 2 end root
f to the power of negative 1 end exponent colon open parentheses negative 2 comma infinity close parentheses rightwards arrow open parentheses 0 comma infinity close parentheses comma space f to the power of negative 1 end exponent left parenthesis x right parenthesis equals square root of x plus 2 end root
f to the power of negative 1 end exponent colon open parentheses 1 divided by 2 comma infinity close parentheses rightwards arrow open parentheses 0 comma infinity close parentheses comma space f to the power of negative 1 end exponent left parenthesis x right parenthesis equals square root of x minus 1 half end root
  Raspunsul este corect.
  Raspunsul este gresit.

Explicatie:

Răspuns corect: A

f bijectivă rightwards double arrow f inversabilă.

f left parenthesis x right parenthesis equals y comma space x element of open parentheses 0 comma infinity close parentheses comma space y element of open parentheses 2 comma infinity close parentheses x squared plus 2 equals y left right double arrow x squared plus 2 minus y equals 0 rightwards double arrow open parentheses for all close parentheses space y element of open parentheses 2 comma infinity close parentheses comma space open parentheses there exists close parentheses space x equals plus-or-minus square root of y minus 2 end root

Dar space x element of open parentheses 0 comma infinity close parentheses space rightwards double arrow x equals square root of y minus 2 end root

rightwards double arrow f to the power of negative 1 end exponent colon open parentheses 2 comma infinity close parentheses rightwards arrow open parentheses 0 comma infinity close parentheses comma space f to the power of negative 1 end exponent left parenthesis x right parenthesis equals square root of x minus 2 end root.