Question

Find the equation of the tangent to the curve y = (2x – 1)e2(1 –x) at the point of its maximum.

Answer 1

Given curve is

y = (2x – 1) e2(1 –x)

⇒ dy/dx = 2e2(1 –x) + (2x – 1) e2(1 –x) + (–2)

For max or min, dy/dx = 0

⇒ 2e2(1 –x) + (2x – 1) e2(1–x) (– 2) = 0

⇒ 1 –(2x – 1) = 0

⇒ 2 – 2x = 0

⇒ x = 1

Clearly, slope = m = 0

when x = 1, then y = 1

Hence, the equation of the tangent is

y – 1 = m(x – 1)

⇒ y – 1 = 0 .(x – 1)

⇒ y – 1 = 0