Matematica - Ecuatii si inecuatii. Functii

Ecuatii si inecuatii. Functii - Intrebarea 6

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Să se determine punctele de intersecție ale graficului funcției $f colon straight real numbers rightwards arrow straight real numbers comma space f left parenthesis x right parenthesis equals 2 to the power of x plus 1 end exponent minus 1$ cu axele de coordonate.

G subscript f intersection O x equals open curly brackets left parenthesis negative 1 semicolon 0 right parenthesis close curly brackets comma space G subscript f intersection O y equals open curly brackets left parenthesis 0 semicolon 1 right parenthesis close curly brackets

G subscript f intersection O x equals open curly brackets left parenthesis 1 semicolon 0 right parenthesis close curly brackets comma space G subscript f intersection O y equals open curly brackets left parenthesis 0 semicolon minus 1 right parenthesis close curly brackets

G subscript f intersection O x equals open curly brackets left parenthesis 1 semicolon 0 right parenthesis close curly brackets comma space G subscript f intersection O y equals open curly brackets left parenthesis 0 semicolon 1 right parenthesis close curly brackets

Raspunsul este corect.

Raspunsul este gresit.

Explicatie:

Răspuns corect: A

$G subscript f intersection O x equals open curly brackets A left parenthesis x semicolon 0 right parenthesis close curly brackets comma space space space space space space space space space space space space space space space space space f left parenthesis x right parenthesis equals 0 left right double arrow 2 to the power of x plus 1 end exponent minus 1 equals 0 left right double arrow x equals negative 1 rightwards double arrow A left parenthesis negative 1 semicolon 0 right parenthesis G subscript f intersection O y equals open curly brackets B left parenthesis 0 semicolon f left parenthesis 0 right parenthesis right parenthesis close curly brackets space space space space space space space space space space space space space space space space f left parenthesis 0 right parenthesis equals 1 rightwards double arrow B left parenthesis 0 semicolon 1 right parenthesis space space space space$