g. On one job, the builder adjusts his template so that \( h(x)=2.2+1.5 \times 3^{-k x} \) where \( x \in[0,2] \). There is a specific requirement on this job that the minimum height of the verandah roof, where it is attached to the posts, must be greater than 2.3 metres and less than 2.4 metres. Find the possible values of \( k \) correct to three decimal places.

To find the possible values of \( k \) such that the minimum height of the verandah roof is between 2.3 and 2.4 meters, we need to analyze the function \( h(x) = 2.2 + 1.5 \times 3^{-kx} \) over the interval \( x \in [0, 2] \). The function \( h(x) \) is a decreasing exponential function because the base of the exponent is between 0 and 1 (specifically \( 3^{-kx} \)). This means that as \( x \) increases, \( h(x) \) decreases. Therefore, the minimum value of \( h(x) \) will occur at the upper bound of the interval, \( x = 2 \). We want to find the values of \( k \) such that: \[ 2.3 < h(2) < 2.4 \] Let's calculate \( h(2) \) and then solve for \( k \). \[ h(2) = 2.2 + 1.5 \times 3^{-2k} \] Now, we set up two inequalities: 1. \( 2.3 < 2.2 + 1.5 \times 3^{-2k} \) 2. \( 2.2 + 1.5 \times 3^{-2k} < 2.4 \) Let's solve each inequality separately. For the first inequality: \[ 2.3 < 2.2 + 1.5 \times 3^{-2k} \] \[ 0.1 < 1.5 \times 3^{-2k} \] \[ \frac{0.1}{1.5} < 3^{-2k} \] \[ \frac{1}{15} < 3^{-2k} \] Taking the natural logarithm of both sides: \[ \ln\left(\frac{1}{15}\right) < -2k \ln(3) \] \[ -\ln(15) < -2k \ln(3) \] \[ \frac{\ln(15)}{2 \ln(3)} > k \] For the second inequality: \[ 2.2 + 1.5 \times 3^{-2k} < 2.4 \] \[ 1.5 \times 3^{-2k} < 0.2 \] \[ 3^{-2k} < \frac{0.2}{1.5} \] \[ 3^{-2k} < \frac{2}{15} \] Taking the natural logarithm of both sides: \[ \ln\left(3^{-2k}\right) < \ln\left(\frac{2}{15}\right) \] \[ -2k \ln(3) < \ln(2) - \ln(15) \] \[ \frac{\ln(2) - \ln(15)}{2 \ln(3)} > k \] Now we have two inequalities for \( k \): \[ k < \frac{\ln(15)}{2 \ln(3)} \] \[ k > \frac{\ln(2) - \ln(15)}{2 \ln(3)} \] Let's calculate the values of \( k \) using a calculator: \[ k < \frac{\ln(15)}{2 \ln(3)} \approx \frac{2.7080503}{2 \times 1.0986123} \approx \frac{2.7080503}{2.1972246} \approx 1.237 \] \[ k > \frac{\ln(2) - \ln(15)}{2 \ln(3)} \approx \frac{0.6931472 - 2.7080503}{2 \times 1.0986123} \approx \frac{-2.0149031}{2.1972246} \approx -0.918 \] Therefore, the possible values of \( k \) are such that: \[ -0.918 < k < 1.237 \] Rounded to three decimal places, the possible values of \( k \) are: \[ -0.918 < k < 1.237 \]