Find a number in the closed interval [1/2,3/2] such that the sum of the number and its reciprocal isa.)as small as possibleb.)as large as possible

Let's assume that number as x

so,

the sum of the number and its reciprocal is

Firstly, we will find derivative

now, we can set it to 0

and then we can solve for x

Since, only x=1 lies on [1/2,3/2]

so, we will consider only x=1

now, we can plug end values and x=1 into S

At x=1/2:

we can plug x=1/2

At x=1:

we can plug x=1

At x=3/2:

we can plug x=3/2

(a)

Smallest value:

The smallest value among them

So, a number is 1..........Answer

(b)

Largest value:

The largest value among them

So, a number is ........Answer

you answer is a

step-by-step explanation:

step-by-step explanation: