Question

The maximum and minimum value of y=x+1/x in interval [1/3,4/3]​

Answer 1

Answer :

• y (max.) = 3⅓

• y (min.) = 2

Solution :

• Given : y = x + 1/x
• To find : maximum and minimum value of y in the interval [ 1/3 , 4/3 ] .

We have ;

y = x + 1/x

Now ,

Differentiating both sides with respect to x , we get ;

=> dy/dx = d(x + 1/x)/dx

=> dy/dx = dx/dx + d(1/x)/dx

=> dy/dx = 1 - 1/x²

We know that ,

When y is minimum or maximum , dy/dx must be zero .

Thus ,

=> dy/dx = 0

=> 1 - 1/x² = 0

=> 1 = 1/x²

=> x² = 1

=> x = √1

=> x = ±1

Now ,

Let's find the values of y in the interval [1/3 , 4/3] for x = 1/3 , 1 , 4/3 .

{ Note : We would not find the value of y at x = -1 as -1 ∉ [1/3 , 4/3] . }

★ At x = 1/3

=> y = x + 1/x

=> y = 1/3 + 3

=> y = (1 + 9)/3

=> y = 10/3

=> y = 3 ⅓

★ At x = 1

=> y = x + 1/x

=> y = 1 + 1

=> y = 2

★ At x = 4/3

=> y = x + 1/x

=> y = 4/3 + 3/4

=> y = (16 + 9)/12

=> y = 25/12

=> y = 2 1⁄12

Clearly ,

• At x = ⅓ , y (max.) = 3 ⅓

• At x = 1 , y (min.) = 2

Answer 2

This is the correct answer .