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Mathematics 7 Online
OpenStudy (dls):

Find f(3)

OpenStudy (dls):

\[\LARGE f(2x)=f'(x).f"(x)\]

OpenStudy (dls):

I replaced x with 3/2

OpenStudy (dls):

\[\LARGE f(3)=f'(\frac{3}{2}) \times f"(\frac{3}{2})\] m stuck here

OpenStudy (dls):

@ganeshie8

OpenStudy (fifciol):

substite for \[f(x)=e^{ax}\]

OpenStudy (fifciol):

find what the f(2x) is than what f'(x) is and f''(x)

OpenStudy (fifciol):

put that in equation and solve for a

OpenStudy (fifciol):

then substitute that a and 3 instead of x and you should come up with the right answer

OpenStudy (dls):

why e^ax?

OpenStudy (fifciol):

i predict that this function could satisfy my equation

OpenStudy (fifciol):

What value did you find?

OpenStudy (dls):

i cudn't find it

OpenStudy (dls):

not getting the correct answer,confused

OpenStudy (fifciol):

well if f(x)=e^ax what is f(2x)?

OpenStudy (dls):

e^2ax

OpenStudy (fifciol):

yes, what about the first and second derivative of f(x)?

OpenStudy (dls):

2e^2ax and 4e^2ax

OpenStudy (fifciol):

a is also constant, but you tried to calculate derivatives of e^2ax but in equation you have to take derivatives of f(x) so these are derivatives of e^ax

OpenStudy (dls):

yes

OpenStudy (fifciol):

so first derivative is a*e^ax and second is a^2*e^ax. Substitute them to equation

OpenStudy (fifciol):

and solve for a

OpenStudy (dls):

I think u missed 2 and 4

OpenStudy (fifciol):

no, we take derivative of \[e^{ax}\]

OpenStudy (dls):

\[\LARGE e^{2ax}=ae^{ax} \times ae^{ax} \]

OpenStudy (fifciol):

\[e^{ax}=ae^{ax}*a^2e^{ax}=a^3*e^{ax+ax}=a^3*e^{2ax} \rightarrow a=1\]

OpenStudy (dls):

im sorry,a=1?

OpenStudy (dls):

yeah..

OpenStudy (fifciol):

so your function f(x) is simply e^x

OpenStudy (dls):

so f(x)=e^x only

OpenStudy (fifciol):

so f(3) =?

OpenStudy (dls):

e cube? :/

OpenStudy (fifciol):

right :)

OpenStudy (dls):

Options: A)4 B)12 C)15 D)None of these

OpenStudy (fifciol):

I think none of these

OpenStudy (dls):

it is 12 ....

OpenStudy (fifciol):

how come?

OpenStudy (dls):

dont know

OpenStudy (calculusxy):

Just insert the 3 to the places in which the x belongs.

OpenStudy (dls):

I guess someone didnt read the question

OpenStudy (calculusxy):

So the confusing part can be when it shows the f(2x). Since the rule is to insert the 3 to x, what you just need to do is multiply the 2 and 3 which will evaluate to 6 making it f(6)

OpenStudy (dls):

read my first reply

OpenStudy (calculusxy):

What do you mean?

OpenStudy (dls):

x->3/2

OpenStudy (calculusxy):

So you're saying that you should input 3/2 to x?

OpenStudy (dls):

yes

OpenStudy (calculusxy):

But that's not how I would evaluate the problem because it is the same thing as 2(3)

OpenStudy (calculusxy):

What I mean to say is that you have to multiply the 2 and 3 since the variable x is right next to 2.

OpenStudy (calculusxy):

That's would be my first cue towards solving this procedure.

OpenStudy (anonymous):

there are many infinitely solutions for the functional equation\[f(2x)=f'(x) f''(x)\]i'd probably say the problem must be something else, \(f\) will be a polynomial i think through ur options

OpenStudy (anonymous):

i got \(12\) with considering \(f\) as a polynomial :)

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