Can you please help me in this taylor polynomial problem?!?

Question

A function f has derivatives of all orders for all real numbers, x. Assume f(3)= −4, f ′(3) =7, f "(3) = 8 and the third derivative of f at x = 3 is −9.

a, Write the 3rd degree Taylor polynomial for f about x = 3 and use it to approximate f (2.5).
b. The fourth derivative of f satisfies the inequality f(4) (x) ≤ 3 for all x in the interval [ 2.5, 3]. Use the LaGrange error bound on the approximation to f (2.5) found in part (a) to explain why f(2.5) cannot equal −8.
c. Write a fourth degree Taylor polynomial P(x), for g(x) = f (x2 + 3) about x = 0. Use P(x) to explain why g must have a relative minimum at x = 0.

Scarlet Manuka · Answer

(a) T(x) = f(3) + f'(3) (x - 3) + f''(3) (x - 3)² / 2! + f'''(3) (x - 3)³ / 3!
= -4 + 7 (x - 3) + 4 (x - 3)² - (3/2) (x - 3)³.
So f(2.5) ≈ T(2.5)
= -4 + 7 (-0.5) + 4 (-0.5)² - (3/2) (-0.5)³
= -6.3125

(b) We know the error will be of form e = f^(4) (x) (-0.5)^4 / 4!  for some x between 2.5 and 3, hence 
|e| ≤ 3 (-0.5)^4 / 24 = 0.0078 (4 d.p.). Hence the true value of f(2.5) must be strictly between -6.3204 and -6.3046.

(c) G(x) = f(x² + 3). For x near 0, x² + 3 will be near 3, so we can use the Taylor expansion of f(x) from before. Wherever we had (x - 3), we can replace is by ((x²+3) - 3) = x². So we get
T(x² + 3) = -4 + 7 (x²) + 4 (x²)² - (3/2) (x²)³
We only want a fourth degree polynomial here, so we drop the last term to get
P(x) = -4 + 7x² + 4x^4.
Near x = 0, g(x) will behave like -4 + 7x², and since 7x² has a minimum at x = 0, so will g(x).

Incidentally, we can also prove this from the chain rule:
g'(x) = f'(x² + 3) . 2x, so g'(0) = f'(3) (0) = 0
and g''(x) = f''(x² + 3) (2x)² + f'(x² + 3) (2)
so g''(0) = f''(3) (0) + f'(3) (2) = 14. 
Since g'(0) = 0 and g''(0) > 0, g(x) has a minimum at x = 0.

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Can you please help me in this taylor polynomial problem?!?

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