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# simple p2 question watch

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

find the range of f
2. (Original post by Bebop)
f(x)=x^2 - 2x + 3 x is a set of real numbers, 0=<x>=4

find the range of f
where is minimum?

2x-2 = 0
x=1
y=1-2+3=2
f>=2
3. (Original post by 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
4. (Original post by lexazver203)
what is the general method of finding a range of f never realy understood how it is done
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.
5. why not just draw a graph?
6. (Original post by TheWolf)
why not just draw a graph?
I agree, however when you get more complex functions drawing them can be really hard!
7. (Original post by Katie Heskins)
I agree, however when you get more complex functions drawing them can be really hard!
yea i guess but how hard are p2 functions? oh and is anyone using graphic calculators?
8. (Original post by 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.
and you get minimum and maximum on what type of graphs? quadratic cubic?
9. (Original post by lexazver203)
and you get minimum and maximum on what type of graphs? quadratic cubic?
any graphs
10. (Original post by TheWolf)
any graphs
that have maximum minimum points that is
11. (Original post by TheWolf)
any graphs
i mean both min and max at the same time
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.
13. (Original post by lexazver203)
i mean both min and max at the same time
functions of order greater than 2. (ie cubics, quartics, upwards) but they won't always have max AND mins
14. (Original post by lexazver203)
what is the general method of finding a range of f never realy understood how it is done
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
========
15. (Original post by 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
========
how do you know f_max is when x=4 is says that x>=4
16. (Original post by lexazver203)
how do you know f_max is when x=4 is says that x>=4
the range of x is given as,

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

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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