Find the maximum value of the function f(x) = –x^2 + 3x + 2.

F( x) = 1/(x^ 2 - 3x + 2)

Given function should be maximum when (x^ 2 - 3x + 2) will be minimum.

Let , y = (x^ 2 - 3x + 2)

Differentiate with respect to x, we get

dy/dx = 2x - 3

For maximum or minimum we will put dy/dx = 0

=> x= 3/2

Now, we find the second derivative.

=> d^2y/dx^2 = 2 (+ ve)

That means (x^ 2 - 3x + 2) is minimum.

Therefore f(x) = 1/(x^ 2 - 3x + 2) is maximum.

=> Maximum value at 3/2 is

=> 1/(9/4 - 9/2 + 2)

=> 4/(9 - 18 + 8)

=> - 4

Hence, the maximum value of the function is - 4