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Suponha que você invista \(\$1000\) em uma conta por \(5\) anos (\(60\) meses), com uma taxa de
juros anual de \(r\), capitalizada mensalmente. A quantia de dinheiro na sua conta após \(5\)
anos será \[ A(r)=1000(1+r/12)^{60} \] Use o Teorema do Valor Intermediário para mostrar que existe um valor de
\(r\) no intervalo \((0,0.1)\), isto é, uma taxa de juros entre \(0\%\) e \(10\%\), que permitirá que você alcance sua
meta de economizar \(\$1500\) em \(5\) anos.
Vamos resolver o problema em etapas.
ETAPA 1
A função \(A\) é contínua e seu domínio é \[ \text {Domínio de }A=[\class{xm-answer}{\cssId{xm-answer-1}{\phantom{\rule{7em}{1.5em}}}},\class{xm-answer}{\cssId{xm-answer-2}{\phantom{\rule{7em}{1.5em}}}}] \] ETAPA 2
Calcule os valores \[ A(0)=\class{xm-answer}{\cssId{xm-answer-3}{\phantom{\rule{7em}{1.5em}}}}, A(\class{xm-answer}{\cssId{xm-answer-4}{\phantom{\rule{7em}{1.5em}}}})=1645.31. \]
Seja \(L=1500\). Selecione todas as afirmações a seguir que estão corretas.
Como \(A(0)< L<A(0.1)\), o TVI implica que existe um número \(r\) em \(\left (0,1500\right )\) tal que \(A(r)=L\). Como \(A(0)< L<A(0.1)\), o TVI
implica que existe um número \(r\) em \(\left (0,0.1\right )\) tal que \(A(r)=L\). Como \(0< L< 0.1\), o TVI implica que existe um
número \(r\) em \(\left (0,0.1\right )\) tal que \(A(r)=L\). Como \(1000< L< 1645.31\), o TVI implica que existe um número \(r\) em \(\left (0,0.1\right )\) tal que \(A(r)=L\).
Quando aplicamos o Teorema do Valor Intermediário a \(A\) no intervalo \(\left [0,0.1\right ]\) com \(L=1500\), temos
garantida a existência de pelo menos um ponto \(r\) tal que \[ A(r)=\class{xm-answer}{\cssId{xm-answer-5}{\phantom{\rule{7em}{1.5em}}}} \]