Verified answer

y(x) = e^(x+2) - e^(2x)

y'(x) = e^(x+2) - 2e^(2x)

y''(x) = e^(x+2) - 4e^(2x)

Critical point when y'(x) = 0

e^(x+2) - 2e^(2x) = 0

e^(x+2) = 2e^(2x)

ln[e^(x+2)] = ln[2e^(2x)]

x+2 = 2x + ln (2)

2x - x = 2 - ln(2)

x = 2 - ln(2)

x ≈ 1.307

y(2 - ln(2)) ≈ 10.237

Critical point at (1.307, 10.237)

y''(x) = e^(x+2) - 4e^(2x)

y''(2 - ln(2)) = 1.735E-18

y''(x) > 0, so critical point at (1.307, 10.237) is a local maximum.