Question

Absolute minimum value of f(x) = (x − 2)² + 5 in the interval [−3, 2] is ______.

पर्याय

• −3

• 2

• 5

• 30

Answer 1

Absolute minimum value of f(x) = (x − 2)² + 5 in the interval [−3, 2] is 5.

Explanation:

A critical point occurs where the derivative f′(x) = 0.

f(x) = (x − 2)² + 5

f′(x) = 2(x − 2)(1) + 0

= 2(x − 2)

Setting f′(x) = 0

2(x − 2) = 0

x = 2

The value x = 2 is within our closed interval [−3, 2].

Evaluate the function at key points

We need to check the values at the endpoints of the interval: x = −3 and x = 2.

At the left endpoint x = −3:

f(−3) = (−3 − 2)² + 5

= (−5)² + 5

= 25 + 5

= 30

At the right endpoint x = 2:

f(2) = (2 − 2)² + 5

= 0² + 5

= 5

Compare values for the absolute minimum

f(−3) = 30

f(2) = 5

The smallest value among these is 5.