Question

Find the maximum and minimum value of 2x^3-15x^2+36x+1 in [1,5]

Answer 1

Answer:

The maximum is 56 at x = 5.

The minimum is 24 at x = 1.

Step-by-step explanation:

The extrema (maximum and minimum) must occur at either:

• critical point where the derivative is zero; OR
• end point of the interval.

For f(x) = 2x³ - 15x² + 36x + 1,

df/dx = 0

=> 6x² - 30x + 36 = 0

=> x² - 5x + 6 = 0

=> ( x - 2 ) ( x - 3 ) = 0

=> x = 2 or x = 3

The end points of the interval are x = 1 and x = 5.

As the extrema must be among the locations x=1, 2, 3 and 5, we just check the value of the function at these places.

f(1) = 2 - 15 + 36 + 1 = 24

f(2) = 16 - 60 + 72 + 1 = 29

f(3) = 54 - 135 + 108 + 1 = 28

f(5) = 250 - 375 + 180 + 1 = 56

Therefore the maximum is 56 at x = 5.

The minimum is 24 at x = 1.

Answer 2

Step-by-step explanation:

hope it helps you follow 5