What is the minimum value of function y = x2 in the interval [1, 6] ?

Given

y = x²

Differentiate:

dy/dx = 2x

For every x in [1, 6],

2x > 0

Therefore, y = x² is strictly increasing on [1, 6].

For an increasing function on a closed interval, the minimum value occurs at the left endpoint.

So,

x = 1

and

y = 1² = 1

Answer

1 (Option B) ✅