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7. Sejam as funções \( f(x)=2 x+3, g(x)=2-3 x \) e \( h(x)=\frac{4 x-1}{2} \) definidas em \( \mathbb{R} \). Para quais valores de \( x \in \mathbb{R} \), tem-se? \( \begin{array}{lll}\text { a) } f(x) \geq g(x) & \text { b) } g(x)
Ask by Peterson Reese. in Brazil
Mar 09,2025

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a) \( x \geq -\frac{1}{5} \) b) \( x > \frac{1}{2} \) c) Para todo \( x \) real

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Para resolver as desigualdades propostas, vamos analisar cada uma delas separadamente. a) Para \( f(x) \geq g(x) \): Temos \( f(x) = 2x + 3 \) e \( g(x) = 2 - 3x \). Igualando as funções, \( 2x + 3 \geq 2 - 3x \). Resolvendo: \[ 2x + 3x \geq 2 - 3 \\ 5x \geq -1 \\ x \geq -\frac{1}{5} \] Portanto, a desigualdade \( f(x) \geq g(x) \) se mantém para \( x \geq -\frac{1}{5} \). b) Para \( g(x) < h(x) \): Os valores são \( g(x) = 2 - 3x \) e \( h(x) = \frac{4x - 1}{2} \). Igualamos as funções para determinar o intervalo. \[ 2 - 3x < \frac{4x - 1}{2} \] Multiplicando tudo por 2 para eliminar o denominador: \[ 4 - 6x < 4x - 1 \\ 4 + 1 < 4x + 6x \\ 5 < 10x \\ x > \frac{1}{2} \] Assim, \( g(x) < h(x) \) para \( x > \frac{1}{2} \). c) Para \( f(x) \geq h(x) \): Agora, com \( f(x) = 2x + 3 \) e \( h(x) = \frac{4x - 1}{2} \): \[ 2x + 3 \geq \frac{4x - 1}{2} \] Multiplicando por 2: \[ 4x + 6 \geq 4x - 1 \\ 6 \geq -1 \] Como essa desigualdade é sempre verdadeira, \( f(x) \geq h(x) \) para todo \( x \in \mathbb{R} \). Resumindo: - a) \( x \geq -\frac{1}{5} \) - b) \( x > \frac{1}{2} \) - c) Verdadeira para todo \( x \in \mathbb{R} \)

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