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Composite functions C3
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f(x)=ax+b, f(2)=19 and ff(0)=55. Find the possible values of a and b.
So far I worked out ff= (a^2)x + ab
ff(0)= a^2 + ab=55
f(2)= 2a+b=19
I'm not sure what to do next. I tried to solve it simultaneously but I couldn't work it out.
So far I worked out ff= (a^2)x + ab
ff(0)= a^2 + ab=55
f(2)= 2a+b=19
I'm not sure what to do next. I tried to solve it simultaneously but I couldn't work it out.
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#2
(Original post by osayukiigbinoba)
f(x)=ax+b, f(2)=19 and ff(0)=55. Find the possible values of a and b.
So far I worked out ff= (a^2)x + ab
ff(0)= a^2 + ab=55
f(2)= 2a+b=19
I'm not sure what to do next. I tried to solve it simultaneously but I couldn't work it out.
f(x)=ax+b, f(2)=19 and ff(0)=55. Find the possible values of a and b.
So far I worked out ff= (a^2)x + ab
ff(0)= a^2 + ab=55
f(2)= 2a+b=19
I'm not sure what to do next. I tried to solve it simultaneously but I couldn't work it out.
ff(x) = f(f(x))= af(x)+b
= a(ax+b) +b
= a^2x + ab + b
Then set x=0
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#3
Is this for edexcel exam board? If so what past paper question is it? I got two sets of values for A and B
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(Original post by ghostwalker)
Your ff(x) is incorrect.
ff(x) = f(f(x))= af(x)+b
= a(ax+b) +b
= a^2x + ab + b
Then set x=0
Your ff(x) is incorrect.
ff(x) = f(f(x))= af(x)+b
= a(ax+b) +b
= a^2x + ab + b
Then set x=0
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(Original post by JoshBrown97)
Is this for edexcel exam board? If so what past paper question is it? I got two sets of values for A and B
Is this for edexcel exam board? If so what past paper question is it? I got two sets of values for A and B
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#6
(Original post by osayukiigbinoba)
It's from an Ocr textbook and it says the answers are a=4, b=11 or a=4.5, b=10
It's from an Ocr textbook and it says the answers are a=4, b=11 or a=4.5, b=10
Therefore you have 2 equations
2a + b = 19 ----> 1
ab + b = 55 ----> 2
Rearrange to find b. -----> b = 55/(a-1)
Sub the equation we have for b into equation 1. which will turn out to be a quadratic and we factorise for a = 9/2 (4.5) or a = 4. Sub these values back into either equation to locate b. The b values turn out to be b = 10 and b = 11
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(Original post by JoshBrown97)
Yes I got them answers. Firstly I solved f(2) = 2a + b (which equalled 19). Then I solved ff(0). f(0) = b then f(b) (as it is ff(0)) equals 55.
Therefore you have 2 equations
2a + b = 19 ----> 1
ab + b = 55 ----> 2
Rearrange to find b. -----> b = 55/(a-1)
Sub the equation we have for b into equation 1. which will turn out to be a quadratic and we factorise for a = 9/2 (4.5) or a = 4. Sub these values back into either equation to locate b. The b values turn out to be b = 10 and b = 11
Yes I got them answers. Firstly I solved f(2) = 2a + b (which equalled 19). Then I solved ff(0). f(0) = b then f(b) (as it is ff(0)) equals 55.
Therefore you have 2 equations
2a + b = 19 ----> 1
ab + b = 55 ----> 2
Rearrange to find b. -----> b = 55/(a-1)
Sub the equation we have for b into equation 1. which will turn out to be a quadratic and we factorise for a = 9/2 (4.5) or a = 4. Sub these values back into either equation to locate b. The b values turn out to be b = 10 and b = 11
often anymore so I just saw this now. Thank you very much for the help.
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