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Find the range of f...

The function f is defined by:

f(x) = x^2 - 3x + 7 , where all real values for x

Find the range of f. [3]


I don't really understand how to do this...
Reply 1
Avatar for Lou Reed
Lou Reed
10
differentiate it and set it equal to 0
Reply 2
Avatar for Lou Reed
Lou Reed
10
(and then find x)
Reply 3
Avatar for Je3y
Je3y
OP
4
I did, but i got 3/2. The answer is 19/4... It showed a different method in the markscheme.

I differentiated it to get 2x - 3, then rearranged... :s-smilie:
Reply 4
Avatar for Lou Reed
Lou Reed
10
Je3y
I did, but i got 3/2. The answer is 19/4... It showed a different method in the markscheme.

I differentiated it to get 2x - 3, then rearranged... :s-smilie:

really?? hmmm

what method did the mark scheme have?
Reply 5
Avatar for Sver
Sver
f(x) = x^2 - 3x + 7
f(x) = (x-3/2)^2-9/4 + 28/4 [complete the square]
f(x) = (x-3/2)^2+19/4

lowest point (3/2, 19/4), y >/ 19/4


also y = x^2-3x+7
y'= 2x-3
0 = 2x-3
3/2 = x
sub into y to get range.
Reply 6
Avatar for Lou Reed
Lou Reed
10
oh dear lol i feel stupid...obviously range means y values (!) we forgot to sub 3/2 into the original..
Reply 7
Avatar for Je3y
Je3y
OP
4
Sver
f(x) = x^2 - 3x + 7
also y = x^2-3x+7
y'= 2x-3
0 = 2x-3
3/2 = x
sub into y to get range.


Ahh stupid me. I forgot to substitute the x = 3/2 into the equation to find y. doh.
Reply 8
Avatar for Je3y
Je3y
OP
4
Lou Reed
oh dear lol i feel stupid...obviously range means y values (!) we forgot to sub 3/2 into the original..


Haha yeah... I have a C3 exam tomorrow aswell!

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