Can someone help me with this calculus problem

Let P(x) = 7 - 3(x - 4) + 5(x - 4)^2 - 2(x - 4)^3 + 6(x - 4)^4 be the fourth-degree Taylor polynomial for the

function f about 4. Assume f has derivatives of all orders for all real numbers.

(a) Find f(4) and f ' ' ' (4).
(b) Write the second degree Taylor polynomial for f ' about 4 and use it
to approximate f '(4.3).

(c) Write the fourth-degree Taylor polynomial for g(x) = ∫ f(t) dt about 4. (integral from 4 to x)

If anyone knows how to do any of these, I would really appreciate it!
Thank you SO much! =]

we know
f(x+h)=f(x)+hf'(x)+h^2/2!*f''(x)+.... eq 1
now f(x)=f(4+x-4)...eq 2 so taking x=4 and h=(x-4) and comparing your equation
a)we get f(4)=7 f'(4)=-3 f''(4)=5*1*2=10 f'''(4)=-2*1*2*3=-12 f''''(4)=6*1*2*3*4=144 (Answer)

b) From equation 1 and 2
f '(x)=f'(4+x-4)=f '(4) + (x-4)f''(4)+(x-4)^2/2*f'''(4)
Substituting values of f '(4), f ''(4) and f '''(4) from a) we get
f '(x)=-3+10(x-4)-6(x-4)^2...eq 3
now substituting x=4.3 in eq 3 we get f '(4.3)= - 0.54 (Answer)

c) g(x) = ∫ f(t) dt(integral from 4 to x)=∫ f(x) dx (integral from 4 to x) from properties of definite integrals
So f(x)=g'(x)...eq 4
Integrating P(x)(integral from 4 to x) we get g(x)=7x - 3/2(x - 4)^2 + 5/3(x - 4)^3 - 2/4(x - 4)^4 + 6(x - 4)^5+.... - 28....eq 5[because4-4=0 we cancel all terms after 28]
so g(4)=0 by substituting x=4 in eq 5
S0 g(x)=g(4+x-4)=g(4)+g'(4)(x-4)+g''(4)/2(x… + g'''(4)/6(x-4)^3 + g''''(4)/24(x-4)^4
from eq 4 we get
g(x)=0+f(4)(x-4)+f '(4)/2(x-4)^2 + f ''(4)/6(x-4)^3 + f '''(4)/24(x-4)^4
Now substituting values of f(4),f'(4),f''(4) and f'''(4) in g(x)
we get g(x)=7(x-4)-3/2(x-4)^2 + 5/3(x-4)^3 - 1/2(x-4)^4 (Answer)