How do you these two questions?

dy/dx = e⁻²ᵗ (1 − t)

d²y/dx² = -e⁻³ᵗ (3 − 2t)

(1.5, ∞)

Step-by-step explanation:

x = eᵗ

y = te⁻ᵗ

dx/dt = eᵗ

dy/dt = -te⁻ᵗ + e⁻ᵗ

dy/dt = e⁻ᵗ (1 − t)

dy/dx = (dy/dt) / (dx/dt)

dy/dx = e⁻ᵗ (1 − t) / eᵗ

dy/dx = e⁻²ᵗ (1 − t)

d²y/dx² = d(dy/dx)/dt / (dx/dt)

d²y/dx² = (-e⁻²ᵗ − 2(1 − t) e⁻²ᵗ) / eᵗ

d²y/dx² = -e⁻²ᵗ (1 + 2(1 − t)) / eᵗ

d²y/dx² = -e⁻³ᵗ (3 − 2t)

The function is concave up when the second derivative is positive.  -e⁻³ᵗ is always negative, so the second derivative is positive when 3 − 2t is negative.

3 − 2t < 0

3 < 2t

t > 3/2

Or, in interval notation, (1.5, ∞).

Because there's a negative sign, it suggests that you should inverse the expression. After inverting it, the power will be positive, and you can square like normal.