Question

evaluate $y = e^x + 1$ for the following values of $x$. round to the nearest thousandth. $x = -2, y \approx \square$ $x = 1, y \approx \square$ $x = 2, y \approx \square$

Explanation:

Step1: Evaluate for \( x = -2 \)

Substitute \( x = -2 \) into \( y = e^x + 1 \). We know that \( e^{-2}=\frac{1}{e^{2}}\approx\frac{1}{7.389056}\approx0.135335 \). Then \( y = e^{-2}+1\approx0.135335 + 1 = 1.135335 \), round to nearest thousandth: \( 1.135 \).

Step2: Evaluate for \( x = 1 \)

Substitute \( x = 1 \) into \( y = e^x + 1 \). \( e^{1}\approx2.718282 \), so \( y = e^{1}+1\approx2.718282 + 1 = 3.718282 \), round to nearest thousandth: \( 3.718 \).

Step3: Evaluate for \( x = 2 \)

Substitute \( x = 2 \) into \( y = e^x + 1 \). \( e^{2}\approx7.389056 \), so \( y = e^{2}+1\approx7.389056 + 1 = 8.389056 \), round to nearest thousandth: \( 8.389 \).

Answer:

For \( x = -2 \), \( y \approx \boldsymbol{1.135} \); for \( x = 1 \), \( y \approx \boldsymbol{3.718} \); for \( x = 2 \), \( y \approx \boldsymbol{8.389} \)