Question

f(x)=2x³-24x+107 x ∈ [1, 3],Find the maximum and minimum value of the given function.

Answer 1

Dear Student,

Answer: Function has minima at x= 2

Solution:

f(x)=2x³-24x+107 x ∈ [1, 3]

f'(x) =
$6 {x}^{2} - 24$
To find the points equate f'(x) = 0

$6 {x}^{2} - 24 = 0 \\ \\ 6( {x}^{2} - 4) = 0 \\ \\ {x}^{2} = 4 \\ x = + 2 \\ x = - 2$
in closed interval x= -2 doesn't lies,so discard x = -2

now find the value of function at x = 1, 2 ,3

f(1) =2(1)³-24(1)+107

= 2-24+107 = 85

f(2) =2(2)³-24(2)+107

= 16-48+107 = 75

f(3) =2(3)³-24(3)+107

= 54-72+107 = 89

So, the function has maximum value in close interval at x= 3, Maximum value= 89.

minimum value at x= 2, minimum value = 75


Hope it helps you

Answer 2

Let f (x) = 2x³ – 24x + 107, x ∈ [1, 3]
⇒ f’(x) = 6x² – 24
=6(x² - 4)
Now, f’(x) = 0
⇒ 6(x² - 4) = 0
⇒ x² = 4
⇒ x = ± 2
Therefore, we will only consider the interval [1,3]
Now, we evaluate the value of f at critical point x = 2 $\in$ [1,3] and at end points of the interval [1,3].
f(2) = 2(2)² – 24(2) + 107
= 2(8) – 24(2) + 107
= 75

f(1) = 2(1)³ – 24(1)+ 107
= 2 – 24 +107
= 85

f(3) = 2(3)³ – 24(3) + 107
= 2(27) -24(3) +107
= 89

maximum value of function is 89 at x = 3 and minimum value of function is 75 at x = 2