# Image of a function.Determine the image of the function f(x)=x^2-6x+5 for x>=4.

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to find the image of the function, hence, you need to evaluate the range of the given function, such that:

`x^2 - 6x + 5 = y => x^2 - 6x = y - 5`

You need to complete the square to the left using the following formula, such that:

`a^2 - 2ab + b^2 = (a - b)^2`

Reasoning by analogy, yields:

`x^2 - 6x + 9 = y - 5 + 9 => (x - 3)^2 = y + 4 => x - 3 = +-sqrt(y + 4) => x = 3 +- sqrt(y + 4)`

Since the problem provides the information that `x >= 4` yields:

`3 +- sqrt(y + 4) >= 4 => +- sqrt(y + 4) >= 4 - 3 => +- sqrt(y + 4) >= 1`

`sqrt(y + 4) >= 1 => y + 4 >= 1 => y >= 1 - 4 => y = f(x) >= -3`

`-sqrt(y + 4) >= 1 => sqrt(y + 4) <= -1 ` invalid

Hence, evaluating the range of the given function, under the given condition, yields `f(x) >= -3.`

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll put y = x^2-6x+5

We'll subtract y both sides:

x^2 - 6x + 5 - y = 0

We'll determine the roots of the equtaion with quadratic formula:

x1 = [6 + sqrt(36 - 4(5-y))]/2

x2 = [6 - sqrt(36 - 4(5-y))]/2

But, from enunciation, we know that x>=4

[6 + sqrt(36 - 4(5-y))]/2 >=4

We'll multiply by 2:

6 + sqrt(36 - 4(5-y) >= 8

We'll subtract 6 both sides:

sqrt(36 - 4(5-y)) >= 2

We'll raise to square both sides:

36 - 4(5-y) >= 4

We'll subtract 36 both sides:

- 4(5-y) >= 4 - 36

- 4(5-y) >= -32

We'll divide by -4:

5-y =< 8

y >= -3

The image of the function belongs to the range [-3 , +infinite), for x >=4.