If f(x) = x^2 -tx +12 , and the minimum value is 11, find the value of t.

Given the graph of the curve f(x) = x^2-tx
+12

We need to find the value of t such that 11 is the minimum
values for the function.

First let us determine the first
derivative.

==> f'(x) = 2x -t =
0

==> 2x = t

==> x =
t/2

Then the function has a minimum value when x =
t/2

==> f(t/2) = 11

==>
(t/2)^2 -t*t/2 + 12 = 11

==> t^2/4 - t^2/2 + 12 =
11

==> (t^2-2t^2)/4 =
-1

==> -t^2/4 = -1

==>
-t^2 = -4

==> t^2 = 4

==>
t= +- 2

Then we have two values for t
:

==> t= { -2,
2}