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

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The function f(x) has an extreme value at f'(x)

f(x) = x^2 - tx + 12

f'(x) = 2x - t

2x - t = 0

=> x = t/2

f(t/2) = (t^2 / 4) - t^2/2 + 12 = 11

=> - t^2 / 4 = -1

=> t^2 = 4

=> t = 2 or t = -2

**The values of t for which the minimum value of x^2 - tx + 12 is 11 are t = 2 and t = -2**

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}**