How do you calculate f''(1) and f'''(1) given #f(x)= 3- 6(x-1) + 4/(2!) (x-1)^2 - 5/(3!) (x-1)^3 + 1/(4!) (x-1)^4#?

Answer 1

The Taylor series for #f(x)# centered at #a# is:

#f(x) = f(a)+f'(a)(x-a)+ (f''(a))/(2!)(x-a)^2+(f'''(a))/(3!)(x-a)^3+ * * * #

So the Taylor series for #f(x)# centered at #1# is:
#f(x) = f(1)+f'(1)(x-1)+ (f''(1))/(2!)(x-1)^2+(f'''(1))/(3!)(x-1)^3+ * * * #
Comparing the given function with the general expression for the Taylor series for #f# centered at #1#, we see that
#f(1) = 3#, and #f'(1) = -6#, and #f''(1) = 4#, and #f'''(1) = -5#, and finallya #f^((4))(x) = 1#.

So

#f''(1) = 4#, and #f'''(1) = -5#.

If you prefer , you could differentiate to get

#f(x)= 3- 6(x-1) + 4/(2!) (x-1)^2 - 5/(3!) (x-1)^3 + 1/(4!) (x-1)^4#
#f'(x) = -6+4(x-1)-5/(2!)(x-1)^2+1/(3!)(x-1)^3#
#f(x)= 4 - 5 (x-1) + 1/(2!) (x-1)^2#

So

#f''(1) = 4#, and #f'''(1) = -5#.
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Answer 2

To calculate ( f''(1) ) and ( f'''(1) ), we need to find the second and third derivatives of the function ( f(x) ), and then evaluate them at ( x = 1 ).

  1. Find ( f'(x) ): [ f'(x) = -6 + 4 \cdot \frac{2!}{2!}(x - 1)^1 - 5 \cdot \frac{3!}{3!}(x - 1)^2 + 1 \cdot \frac{4!}{4!}(x - 1)^3 ] Simplify: [ f'(x) = -6 + 4(x - 1) - 5(x - 1)^2 + (x - 1)^3 ]

  2. Find ( f''(x) ): [ f''(x) = 4 - 10(x - 1) + 3(x - 1)^2 ]

  3. Find ( f'''(x) ): [ f'''(x) = -10 + 6(x - 1) ]

  4. Evaluate at ( x = 1 ): [ f''(1) = 4 - 10(1 - 1) + 3(1 - 1)^2 = 4 ] [ f'''(1) = -10 + 6(1 - 1) = -10 ]

Therefore, ( f''(1) = 4 ) and ( f'''(1) = -10 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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