Verify if f(x)<3, given the function f(x)=x^2-4x+6.

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f(x) = (x-2)^2 -4 + 6 =>

(x-2)^2 + 2 = positve + 2.

Therefore f(x) >=2 for all x.

Therefore the given inequality,

f(x) < 3 is true if x^2-4x+6 <3 =>

x^2-4x+3<0==>

(x-3)(x-1) <0 =>

1<x<3 .

Therefore, f(x)<3 for x for which 1<x<3.

You need to solve the inequality `x^2 - 4x + 6 lt 3` .

Subtracting 3 both sides yields:

`x^2 - 4x + 6 - 3 lt 3 - 3 =gt x^2 - 4x + 3 lt 0`

`` You need to find the values of x such that `x^2 - 4x + 3 lt 0` .

Find what are the zeroes of `x^2 - 4x + 3` .

`x^2 - 4x + 3 = 0`

Using the quadratic formula yields:

`x_(1,2) = (-b+-sqrt(b^2 - 4ac))/(2a)`

`` `x_(1,2) = (-(-4)+-sqrt(16 - 12))/2`

`x_(1,2) = (4+-2)/2 =gt x_1 = 3; x_2= 1`

Take a value for x<1.

`x = 0 =gt 0^2 - 4*0 + 3 = 3 gt 0`

Take a value for x>3.

`x = 4 =gt 4^2 - 4*4+ 3 = 3 gt 0`

Take a value for x such that 1<x<3.

`x = 2 = gt 2^2 - 4*2 + 3 = 4-8+3 = -1lt 0`

**All the values of x `in` (1;3) prove that `x^2 - 4x + 3 lt 3` .**

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