# simple p2 question

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f(x)=x^2 - 2x + 3 x is a set of real numbers, 0=<x>=4

find the range of f

find the range of f

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#2

(Original post by

f(x)=x^2 - 2x + 3 x is a set of real numbers, 0=<x>=4

find the range of f

**Bebop**)f(x)=x^2 - 2x + 3 x is a set of real numbers, 0=<x>=4

find the range of f

2x-2 = 0

x=1

y=1-2+3=2

f>=2

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#3

(Original post by

where is minimum?

2x-2 = 0

x=1

y=1-2+3=2

f>=2

**kikzen**)where is minimum?

2x-2 = 0

x=1

y=1-2+3=2

f>=2

what is the general method of finding a range of f never realy understood how it is done

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#4

(Original post by

what is the general method of finding a range of f never realy understood how it is done

**lexazver203**)what is the general method of finding a range of f never realy understood how it is done

In the example that you gave as kikzen wrote, there's only a minimum; this gave the smallest value of f(x) so every f(x) value must be equal to or above this.

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#6

(Original post by

why not just draw a graph?

**TheWolf**)why not just draw a graph?

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#7

(Original post by

I agree, however when you get more complex functions drawing them can be really hard!

**Katie Heskins**)I agree, however when you get more complex functions drawing them can be really hard!

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#8

(Original post by

Differentiate to find the maximum and/or minimum values, this will give you the largest and smallest values of f(x), hence your range.

In the example that you gave as kikzen wrote, there's only a minimum; this gave the smallest value of f(x) so every f(x) value must be equal to or above this.

**Katie Heskins**)Differentiate to find the maximum and/or minimum values, this will give you the largest and smallest values of f(x), hence your range.

In the example that you gave as kikzen wrote, there's only a minimum; this gave the smallest value of f(x) so every f(x) value must be equal to or above this.

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#9

(Original post by

and you get minimum and maximum on what type of graphs? quadratic cubic?

**lexazver203**)and you get minimum and maximum on what type of graphs? quadratic cubic?

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#12

I just had a thought; sometimes its gonna be pointless finding the max and mins because a graph might not have turning points, but it may asymptotes and therefore you'll have a restricted range then.

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#13

(Original post by

i mean both min and max at the same time

**lexazver203**)i mean both min and max at the same time

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#14

**lexazver203**)

what is the general method of finding a range of f never realy understood how it is done

**for the range given for x**(the domain).

You should try to visualise a "box" containing the curve, with the x-axis having only the x-values of the range of x and the y-axis having only the y-values of the range of y.

You have a quadratic, so it's a parabola. Then you know it's "U-shaped". Since the coefft of x² is positve then the curved bit is at the bottom, and it has a minimum there. So find the value of x for that mimimum and the y-value for that x-value is the low-end of the range of y=f(x).

To get the hi-end for y (i.e. f), just plug in the max value of x.

f= x² - 2x + 3

f' = 2x - 2 = 0 @ x = 1

f_min = f(x=1) = 1 - 2 + 3 = 2

f_min = 2

======

f_max = f(x=4) = 16 - 8 + 3 = 11

f_max = 11

=======

range of f is given by,

2 <= f <= 11

========

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#15

(Original post by

In general, the simplest way is to try and visualise the curve, then find out what the max and min values for f will be (the range of f, or co-domain)

You should try to visualise a "box" containing the curve, with the x-axis having only the x-values of the range of x and the y-axis having only the y-values of the range of y.

You have a quadratic, so it's a parabola. Then you know it's "U-shaped". Since the coefft of x² is positve then the curved bit is at the bottom, and it has a minimum there. So find the value of x for that mimimum and the y-value for that x-value is the low-end of the range of y=f(x).

To get the hi-end for y (i.e. f), just plug in the max value of x.

f= x² - 2x + 3

f' = 2x - 2 = 0 @ x = 1

f_min = f(x=1) = 1 - 2 + 3 = 2

f_min = 2

======

f_max = f(x=4) = 16 - 8 + 3 = 11

f_max = 11

=======

range of f is given by,

2 <= f <= 11

========

**Fermat**)In general, the simplest way is to try and visualise the curve, then find out what the max and min values for f will be (the range of f, or co-domain)

**for the range given for x**(the domain).You should try to visualise a "box" containing the curve, with the x-axis having only the x-values of the range of x and the y-axis having only the y-values of the range of y.

You have a quadratic, so it's a parabola. Then you know it's "U-shaped". Since the coefft of x² is positve then the curved bit is at the bottom, and it has a minimum there. So find the value of x for that mimimum and the y-value for that x-value is the low-end of the range of y=f(x).

To get the hi-end for y (i.e. f), just plug in the max value of x.

f= x² - 2x + 3

f' = 2x - 2 = 0 @ x = 1

f_min = f(x=1) = 1 - 2 + 3 = 2

f_min = 2

======

f_max = f(x=4) = 16 - 8 + 3 = 11

f_max = 11

=======

range of f is given by,

2 <= f <= 11

========

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#16

(Original post by

how do you know f_max is when x=4 is says that x>=4

**lexazver203**)how do you know f_max is when x=4 is says that x>=4

0=<x>=4

which is obviously a typo!

Split the inequality into two parts, viz.

0=<x --- (1

and

x>=4 --- (2

re-arranging (1 gives,

x >= 0

but we already have,

x > = 4

and if x>= 4 then it is also >= 0 so that inequality (1 is superfluous. It looks reasonable to assume that inequality 4) should be x<=4 rather than x>=4, giving

x >= 0

x <= 4

or,

0 <= x <= 4

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