C3 functions Watch

madnutcase434
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#1
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I am really stuck on this questions about functions from C3. Could someone help me please?? The actual bit I cannot do is finding the value of x:

f(x)=ln(3-2x) x<1.5
g(x)=e2x+1
Find the exact value of x for which f-1(x)=g(x).

So far I have got f-1(x)= (3-ex)/2

Then I tried:

(3-ex)/2 = e2x+1
3-ex = 2e2x+2
0=2e2x+ex-1
0=(2ex-1)(ex+1)

So I got x=ln(0.5) but this is apparently wrong. I don't even know if this is the correct way to find x or, if it is, where I have gone wrong! A little help please?
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TenOfThem
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(Original post by madnutcase434)

So I got x=ln(0.5) but this is apparently wrong. I don't even know if this is the correct way to find x or, if it is, where I have gone wrong! A little help please?
Looks ok to me - what do you think you should have
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brianeverit
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(Original post by madnutcase434)
I am really stuck on this questions about functions from C3. Could someone help me please?? The actual bit I cannot do is finding the value of x:

f(x)=ln(3-2x) x<1.5
g(x)=e2x+1
Find the exact value of x for which f-1(x)=g(x).

So far I have got f-1(x)= (3-ex)/2

Then I tried:

(3-ex)/2 = e2x+1
3-ex = 2e2x+2
0=2e2x+ex-1
0=(2ex-1)(ex+1)

So I got x=ln(0.5) but this is apparently wrong. I don't even know if this is the correct way to find x or, if it is, where I have gone wrong! A little help please?
Have you copied the question correctly? I don't understand what the < sign is there for
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Mr M
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(Original post by brianeverit)
Have you copied the question correctly? I don't understand what the < sign is there for
Really? Try setting x >=1.5 and see what happens to f(x).
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madnutcase434
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The answer is apparently x=ln2... and the question definitely said x<1.5. The other answer I got for x was ln(-1) which is not a possible answer. Thanks for your help anyway
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james22
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(Original post by madnutcase434)
The answer is apparently x=ln2... and the question definitely said x<1.5. The other answer I got for x was ln(-1) which is not a possible answer. Thanks for your help anyway
You sure the answer given was not x=-ln(2)?
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brianeverit
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(Original post by Mr M)
Really? Try setting x >=1.5 and see what happens to f(x).
I misread it as t=\ln(3-2x)x rather than the x<1.5 is just a restriction on the value of x.
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