# calculus

posted by
**Meredith**
.

Let f be a function that has derivatives of all orders for all real numbers. Assume f(0)=5, f'(0)=-3, f''(0)=1, and f'''(0)=4.

Write the third-degree Taylor polynomial for h, where h(x) = integral of f(t)dt from 0 to x, about x=0

for this part, I got 5x-3x^2/2+x^3/6.

Then the second part asks, Let h be defined as the part above. Given that f(1)=3, either find the exact value of h(1) or explain why it cannot be determined.

I am not understanding this part. Please help me out.

The Taylor polynomial for f(x) is 5 -3x + (1/2)*1*x^2 + (1/6)*4*x^4

The Taylor polynomial for h(x) is

5x -(3/2)x^2 + (1/6)x^3 + ...

It is not necessary from the f(0) derivatives provided that the value of f(1) be 3. That is additional information. Since the exact funtion has not been specified, the exact value of the integral from 0 to 1 caqnnot be specified either, even when f(0) and f(1) are known.

I agree that this is a very confusing question.