Question

Find a number in the closed interval [1/2,3/2] such that the sum of the number and its reciprocal is

Answer 1

Let's assume that number as x

so,

the sum of the number and its reciprocal is

$S(x)=x+\frac{1}{x}$

Firstly, we will find derivative

$S'(x)=1-\frac{1}{x^2}$

now, we can set it to 0

and then we can solve for x

$S'(x)=1-\frac{1}{x^2}=0$

$x=-1,x=1$

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

$S(\frac{1}{2})=\frac{1}{2} +\frac{1}{\frac{1}{2}}$

$S(\frac{1}{2})=\frac{1}{2} +2$

$S(\frac{1}{2})=\frac{5}{2}$

At x=1:

we can plug x=1

$S(1)=1 +\frac{1}{1}$

$S(1)=1 +1$

$S(1)=2$

At x=3/2:

we can plug x=3/2

$S(\frac{3}{2})=\frac{3}{2} +\frac{1}{\frac{3}{2}}$

$S(\frac{3}{2})=\frac{3}{2} +\frac{2}{3}$

$S(\frac{3}{2})=\frac{13}{6}$

(a)

Smallest value:

The smallest value among them

$S(1)=2$

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

(b)

Largest value:

The largest value among them

$S(\frac{1}{2})=\frac{5}{2}$

So, a number is $\frac{1}{2}$........Answer