Question

Determine the maximum and minimum value of f(x)=4x^2-5x+1 in the interval [-1.5,1.5]?

Answer 1

Answer:

f(x)=2x

3

−15x

2

+36x+1

f

′

(x)=6x

2

−30x+36

Putting f

′

(x)=0

6x

2

−30x+36=0

x

2

−5x+6=0

(x−2)(x−3)=0

x=2 or 3

We are given interval [1,5]

Hence, calculating f(x) at 2, 3, 1, 5,

f(2)=29

f(3)=28

f(1)=24

f(5)=56

Hence, absolute maximum value is 56 at x=5.

Absolute minimum value is 24 at x=1.