Find the derivative of the function using the definition of derivative. \[ G(t)=\frac{1-2 t}{4+t} \] \( G^{\prime}(t)=\frac{-9}{(4+t)^{2}} \) State the domain of the function. (Enter your answer using interval notation.)

To find the derivative of the function \( G(t) = \frac{1 - 2t}{4 + t} \) using the definition of the derivative, we start with the limit definition: \[ G'(t) = \lim_{h \to 0} \frac{G(t + h) - G(t)}{h} \] First, we need to compute \( G(t + h) \): \[ G(t + h) = \frac{1 - 2(t + h)}{4 + (t + h)} = \frac{1 - 2t - 2h}{4 + t + h} \] Now, we can substitute \( G(t + h) \) and \( G(t) \) into the limit: \[ G'(t) = \lim_{h \to 0} \frac{\frac{1 - 2t - 2h}{4 + t + h} - \frac{1 - 2t}{4 + t}}{h} \] To combine the fractions in the numerator, we find a common denominator, which is \( (4 + t + h)(4 + t) \): \[ G'(t) = \lim_{h \to 0} \frac{(1 - 2t - 2h)(4 + t) - (1 - 2t)(4 + t + h)}{h(4 + t + h)(4 + t)} \] Expanding both terms in the numerator: 1. For \( (1 - 2t - 2h)(4 + t) \): \[ = (1 - 2t)(4 + t) - 2h(4 + t) = (4 + t - 8t - 2t^2) - 2h(4 + t) \] \[ = 4 + t - 8t - 2t^2 - 2h(4 + t) = 4 - 7t - 2t^2 - 2h(4 + t) \] 2. For \( (1 - 2t)(4 + t + h) \): \[ = (1 - 2t)(4 + t) + (1 - 2t)h = (4 + t - 8t - 2t^2) + (1 - 2t)h \] \[ = 4 - 7t - 2t^2 + (1 - 2t)h \] Now, substituting these back into the limit: \[ G'(t) = \lim_{h \to 0} \frac{(4 - 7t - 2t^2 - 2h(4 + t)) - (4 - 7t - 2t^2 + (1 - 2t)h)}{h(4 + t + h)(4 + t)} \] Simplifying the numerator: \[ = \lim_{h \to 0} \frac{-2h(4 + t) - (1 - 2t)h}{h(4 + t + h)(4 + t)} \] \[ = \lim_{h \to 0} \frac{h(-2(4 + t) - (1 - 2t))}{h(4 + t + h)(4 + t)} \] Canceling \( h \): \[ = \lim_{h \to 0} \frac{-2(4 + t) - (1 - 2t)}{(4 + t + h)(4 + t)} \] As \( h \to 0 \): \[ = \frac{-2(4 + t) - (1 - 2t)}{(4 + t)(4 + t)} = \frac{-8 - 2t - 1 + 2t}{(4 + t)^2} = \frac{-9}{(4 + t)^2} \] Thus, we have: \[ G'(t) = \frac{-9}{(4 + t)^2} \] Now, for the domain of the function \( G(t) \): The function \( G(t) = \frac{1 - 2t}{4 + t} \) is defined for all \( t \) except where the denominator is zero. Setting the denominator equal to zero: \[ 4 + t = 0 \implies t = -4 \] Thus, the domain of \( G(t) \) is all real numbers except \( t = -4 \). In interval notation, the domain is: \[ (-\infty,