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functions help please :)

Hi

Im stuck on this question...

ive already worked out that:

A[D(x)] = 2(e^2x) +1

and D[A(x)] = e^(2(2x+1))

it then says, using the substitution t=e^2x ,form and solve a quadratic equation in t to find all possible values of x for which A[D(x)] = D[A(x)]

but im not sure how to sub t into the 2nd one, as the powers confuse me..

thanks :smile:
e(2(2x+1))=e4xe2e^{(2(2x+1))} = e^{4x} e^2
Reply 2
Avatar for nuodai
nuodai
17
If you simplify A(D(x)) and D(A(x)) you get:
A(D(x))=2e2x+1[br]D(A(x))=e4x+2=e4xe2=(e2x)2e2\newline A(D(x)) = 2e^{2x} + 1\newline[br]D(A(x)) = e^{4x + 2} = e^{4x}e^2 = (e^{2x})^2e^2

Now you just have to sub in t=e2xt = e^{2x} and let A(D(x))=D(A(x))A(D(x)) = D(A(x))
Reply 3
Avatar for chbutterfly
chbutterfly
OP
2
thank you :smile:

ive got to 2t+1=t^2 e^2, is that right?

then how do i form the quadratic equation?
Reply 4
Avatar for Jake22
Jake22
18
that is a quadratic i.e. e2t2−2t−1=0e^2t^2 - 2t - 1 = 0

Remember that e2e^2 is just a constant (very roughly 7.4)
Reply 5
Avatar for chbutterfly
chbutterfly
OP
2
ive got 7.39t^2 - 2t -1 =0

but i cant get that into 2 brackets, ive tried..
chbutterfly
ive got 7.39t^2 - 2t -1 =0

but i cant get that into 2 brackets, ive tried..

Quadratic formula, then?

Also, don't use 7.39, use e^2. Then leave your answer in terms of e^2 unless you're asked to do otherwise, in which case you should convert right at the end.

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