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ICtION part II VLN Compute and simplify the difference quotient for each function given. a) \( f(x)=2 x-3 \) b) \( g(x)=x^{2}+3 \) c) \( h(x)=2 \)

Ask by Griffiths Morrison. in the United States
Mar 11,2025

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**a) For \( f(x)=2x-3 \):** \[ \frac{f(x+h)-f(x)}{h} = 2 \] **b) For \( g(x)=x^2+3 \):** \[ \frac{g(x+h)-g(x)}{h} = 2x + h \] **c) For \( h(x)=2 \):** \[ \frac{h(x+h)-h(x)}{h} = 0 \]

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To compute and simplify the difference quotient for each function, we use the formula: \[ \frac{f(x+h) - f(x)}{h} \] For \( f(x) = 2x - 3 \): Substituting, we find \( f(x+h) = 2(x+h) - 3 = 2x + 2h - 3 \). Now, \[ \frac{(2x + 2h - 3) - (2x - 3)}{h} = \frac{2h}{h} = 2 \] For \( g(x) = x^{2} + 3 \): Here, \( g(x+h) = (x+h)^{2} + 3 = x^2 + 2xh + h^2 + 3 \). Thus, \[ \frac{(x^2 + 2xh + h^2 + 3) - (x^2 + 3)}{h} = \frac{2xh + h^2}{h} = 2x + h \] For \( h(x) = 2 \): Since \( h \) is a constant function, \( h(x+h) = 2 \). Therefore, \[ \frac{2 - 2}{h} = \frac{0}{h} = 0 \quad (\text{for } h \neq 0) \] So, the difference quotients are: - \( f(x) \rightarrow 2 \) - \( g(x) \rightarrow 2x + h \) - \( h(x) \rightarrow 0 \)

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