Finding k from Moment Generating Function at t=0

[little neutrino]

Homework Statement

If M[X(t)] = k (2 + 3e^t)^4 , what is the value of k

Homework Equations

M[X(t)] = integral ( e^tx * f(x) )dx if X is continuous

The Attempt at a Solution

I tried differentiating both sides to find f(x), but since it is a definite integral from negative infinity to infinity this method doesn't work. Is this approach (trying to find f(x)) correct? If so, how should I proceed from here? Even if I expand (2 + 3e^t)^4 the resulting expression will be very convoluted and hard to work backwards with. Thanks!

 

[axmls]

Hint: what properties must a moment generating function have? i.e. there is a particularly helpful property of moment generating functions that would help here.

 

[little neutrino]

Hint: what properties must a moment generating function have? i.e. there is a particularly helpful property of moment generating functions that would help here.

Hmm, I'm not sure, is it the Taylor series expansion of M(t)? I worked out the first few terms but doesn't seem to help. I'll just get M(t) = k(2+3e^t)^4 = M(0) + M'(0)t + (M''(0)/2!)t^2 +... = 625k + 1500kt + ... (did not expand the rest, because I don't think it's getting me anywhere)
I don't think this is the particularly helpful property you are referring to... Any hints on what the property is? Thanks!

 

[axmls]

I'll expand on the hint a little. Note that for a continuous random variable, the moment generating function is, as you have pointed out, $$M(t)=\int _{-\infty} ^{\infty} e^{tX} f(x) \ dx$$
What, then, does this say about the value of any moment generating function at $t = 0$?

 

[little neutrino]

I'll expand on the hint a little. Note that for a continuous random variable, the moment generating function is, as you have pointed out, $$M(t)=\int _{-\infty} ^{\infty} e^{tX} f(x) \ dx$$
What, then, does this say about the value of any moment generating function at $t = 0$?

Ok I got it! M(0) = integral f(x) = 1
k = 1/625
Thanks for your help!