# InequalityProve that x^2-6x+13>=4 for real x numbers.

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### 2 Answers

You may also use calculus method to prove that the quadratic function `y = x^2-6x+13` is larger than 4, hence, you need to test that y = 4 is the minimum value of the function `y = x^2-6x+13` , such that:

`(dy)/(dx) = (d(x^2-6x+13))/(dx) => (dy)/(dx) = 2x - 6`

You need to remember that solving the equation `(dy)/(dx) = 0` , you may find the critical values of the function, such that:

`2x - 6 = 0 => 2x = 6 => x = 6/2 => x = 3`

Hence, the critical value of the function is x = 3 and you need to test if `f(3) >= 4` , thus, you need to replace 3 for x in equation of teh function, such that:

`f(3) = 3^2-6*3+13 => f(3) = 9 - 18 + 13 => f(3) = 4`

**Hence, testing if the minimum of the function `y = x^2 - 6x + 13` is 4 yields that the inequality `x^2 - 6x + 13 >= 4` holds and y = 4 represents the minimum value of the function **`y = x^2 - 6x + 13.`

We'll have to determine the interval of numbers that make the inequality to hold.

In this case, we'll have to subtract 4 both sides:

x^2 - 6x+13 - 4>=0

x^2 - 6x+ 9 >=0

We note that the expression above is a perfect square:

(x-3)^2 >= 0

The square of the expression x-3 is always positive for any value of x, except x = 3 for the expression is zero.