Maxima and Minima

Maxima and minima is a very important chapter as far as CAT, XAT, SNAP etc exams are concerned. 

1. Find the minimum value of x2−4x+5$x^2-4x+5$.
We know that graph of this equation is concave up.  As you can see from the graph that it won't touch the x-axis so it does not have any real root. But we can find, where this graph attains its minima we can't find maxima.

Differentiating the given function we get 2x - 4.
We equate this expression to zero to find where minima exists.
2x - 4 = 0 and x = 2.
Substituting in the given expression we get 22−4.2+5$2^2-4.2+5$ = 1.

2. Find the maxima value of 2−2x−x2$2-2x-x^2$.
x2$x^2$ coefficient is negative.  As this graph is concave down, it has maxima. We can't find the minimum

Differentiating the given expression we get, - 2 - 2x.  

Equating to zero, we get x = -1

So at x  = -1 it attains maxima which is equal to 2 - 2 (-1)-(−1)2${\left(-1\right)}^2$ = 3

3. y = max (2-2x, x - 3) then find the minimum value of this function.
The given function is a combination of two linear equations. 2 - 2x is a downward sloping line, and x - 3 is a upward sloping line. As y is defined as max of these two equations, y can be represented as the graph noted with red line. That is upto some point in between 1 and 2, it decreases, and starts increasing after that point.  so this graph attains minimum where these two lines intersect.

Equating, 2 - 2x = x - 3 

We get x = 5/ 3

So minimum value can be obtained by substituting x value in any of these linear equations. 2 - 2(5/3) = -4/3.