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
