(MACK-SP) Dada a função definida por f(x) = 3x + 1, calcule : f(235)-f(129)/160
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Resolução da questão, veja:
Vamos primeiramente calcular f (235) e f (129) para depois se apegarmos à fórmula que é dada na questão, observe:
![\mathsf{f(x) = 3x+1}}~\to~\mathsf{f(235)}}}\\\\\\\\\ \mathsf{f(235) = 3~\cdot~235+1}}\\\\\\\ \mathsf{f(235) = 705 + 1}}\\\\\\\\ \large\boxed{\boxed{\boxed{\boxed{\boxed{\mathsf{f(235)= 706.}}}}}}}}}}}}}}}}}}}} \mathsf{f(x) = 3x+1}}~\to~\mathsf{f(235)}}}\\\\\\\\\ \mathsf{f(235) = 3~\cdot~235+1}}\\\\\\\ \mathsf{f(235) = 705 + 1}}\\\\\\\\ \large\boxed{\boxed{\boxed{\boxed{\boxed{\mathsf{f(235)= 706.}}}}}}}}}}}}}}}}}}}}](https://tex.z-dn.net/?f=+%5Cmathsf%7Bf%28x%29+%3D+3x%2B1%7D%7D%7E%5Cto%7E%5Cmathsf%7Bf%28235%29%7D%7D%7D%5C%5C%5C%5C%5C%5C%5C%5C%5C+%5Cmathsf%7Bf%28235%29+%3D+3%7E%5Ccdot%7E235%2B1%7D%7D%5C%5C%5C%5C%5C%5C%5C+%5Cmathsf%7Bf%28235%29+%3D+705+%2B+1%7D%7D%5C%5C%5C%5C%5C%5C%5C%5C+%5Clarge%5Cboxed%7B%5Cboxed%7B%5Cboxed%7B%5Cboxed%7B%5Cboxed%7B%5Cmathsf%7Bf%28235%29%3D+706.%7D%7D%7D%7D%7D%7D%7D%7D%7D%7D%7D%7D%7D%7D%7D%7D%7D%7D%7D%7D+)
Agora vamos determinar f (129), observe:
![\mathsf{f(x) = 3x+1}}~\to~\mathsf{f(129)}}}\\\\\\\\\ \mathsf{f(129) = 3~\cdot~129+1}}\\\\\\\ \mathsf{f(129) = 387 + 1}}\\\\\\\\ \large\boxed{\boxed{\boxed{\boxed{\boxed{\mathsf{f(129)= 388.}}}}}}}}}}}}}}}}}}}} \mathsf{f(x) = 3x+1}}~\to~\mathsf{f(129)}}}\\\\\\\\\ \mathsf{f(129) = 3~\cdot~129+1}}\\\\\\\ \mathsf{f(129) = 387 + 1}}\\\\\\\\ \large\boxed{\boxed{\boxed{\boxed{\boxed{\mathsf{f(129)= 388.}}}}}}}}}}}}}}}}}}}}](https://tex.z-dn.net/?f=+%5Cmathsf%7Bf%28x%29+%3D+3x%2B1%7D%7D%7E%5Cto%7E%5Cmathsf%7Bf%28129%29%7D%7D%7D%5C%5C%5C%5C%5C%5C%5C%5C%5C+%5Cmathsf%7Bf%28129%29+%3D+3%7E%5Ccdot%7E129%2B1%7D%7D%5C%5C%5C%5C%5C%5C%5C+%5Cmathsf%7Bf%28129%29+%3D+387+%2B+1%7D%7D%5C%5C%5C%5C%5C%5C%5C%5C+%5Clarge%5Cboxed%7B%5Cboxed%7B%5Cboxed%7B%5Cboxed%7B%5Cboxed%7B%5Cmathsf%7Bf%28129%29%3D+388.%7D%7D%7D%7D%7D%7D%7D%7D%7D%7D%7D%7D%7D%7D%7D%7D%7D%7D%7D%7D+)
Agora que temos f (235) e f (129) podemos aplicar à fórmula que é dada na questão, veja:
![\mathsf{\dfrac{f(235)-f(129)}{160}}}\\\\\\\\\ \mathsf{\dfrac{706 - 388}{160}}}\\\\\\\\ \mathsf{\dfrac{318}{160}}}}\\\\\\\\ \large\boxed{\boxed{\boxed{\boxed{\mathsf{\approx2.}}}}}}}}}}}}}}}}}}} \mathsf{\dfrac{f(235)-f(129)}{160}}}\\\\\\\\\ \mathsf{\dfrac{706 - 388}{160}}}\\\\\\\\ \mathsf{\dfrac{318}{160}}}}\\\\\\\\ \large\boxed{\boxed{\boxed{\boxed{\mathsf{\approx2.}}}}}}}}}}}}}}}}}}}](https://tex.z-dn.net/?f=+%5Cmathsf%7B%5Cdfrac%7Bf%28235%29-f%28129%29%7D%7B160%7D%7D%7D%5C%5C%5C%5C%5C%5C%5C%5C%5C+%5Cmathsf%7B%5Cdfrac%7B706+-+388%7D%7B160%7D%7D%7D%5C%5C%5C%5C%5C%5C%5C%5C+%5Cmathsf%7B%5Cdfrac%7B318%7D%7B160%7D%7D%7D%7D%5C%5C%5C%5C%5C%5C%5C%5C+%5Clarge%5Cboxed%7B%5Cboxed%7B%5Cboxed%7B%5Cboxed%7B%5Cmathsf%7B%5Capprox2.%7D%7D%7D%7D%7D%7D%7D%7D%7D%7D%7D%7D%7D%7D%7D%7D%7D%7D%7D+)
Espero que te ajude '-'
Vamos primeiramente calcular f (235) e f (129) para depois se apegarmos à fórmula que é dada na questão, observe:
Agora vamos determinar f (129), observe:
Agora que temos f (235) e f (129) podemos aplicar à fórmula que é dada na questão, veja:
Espero que te ajude '-'
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