Final answer:
The first three nonzero terms of the Taylor expansion for e⁴ˣ about x=3 are found by calculating the function and its derivatives at x=3, yielding e¹²[1+4(x-3)+16(x-3)²+...], which corresponds to option C.
Step-by-step explanation:
The first three nonzero terms of the Taylor expansion for e⁴ˣ expanded about the value 3 can be found using the formula for a Taylor series. The general form of a Taylor series for a function f(x) about a point a is:
f(a) + f'(a)(x - a) + ⅓f''(a)(x - a)² + ...
In this case, the function f(x) = e⁴ˣ, and we are expanding about a = 3. We need the first derivative f'(x) = 4e⁴ˣ and the second derivative f''(x) = 16e⁴ˣ. Substituting a = 3 gives us the corresponding terms:
f(3) = e¹², f'(3) = 4e¹², and f''(3) = 16e¹².
Inserting these into the Taylor series formula, we get:
e¹² [1 + 4(x - 3) + ⅓(16e¹²)(x - 3)² + ...]
This simplifies to:
e¹² [1 + 4(x - 3) + 16(x-3)² + ...]
Therefore, the correct answer is C) e¹²[1+4(x-3)+16(x-3)²+...].