Question

Expand 3x^3 - 2x^2+ x-4 in powers of (x+2)
using Taylors theorem

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Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

Expansion of 3x³ - 2x²+ x - 4 = 3(x +2)³  - 20(x+2)²  + 45(x +2) - 30

Step-by-step explanation:

Given:  

3x³ - 2x²+ x  + 4

To Find:

Expand in powers of (x+2)  using Taylors theorem

Solution:

Now, we are given the following function:

f(x) = 3x³ - 2x²+ x + 4

We calculate the first derivative of the given function and get the following:

f"(x) = 9x² - 4x + 1

We then calculate the second derivative of the above function and get the following:

f''(x)  = 18x  - 4

We calculate the third derivative and get the following:

f'''(x)   = 18

Now, we shall use Taylors theorem to expand the given function f(x). To do that, we shall use the following formula:

f(x)  = f(c)  + f'(c)(x - c)  + f''(c)(x - c)² / 2! + f'''(c)(x - c)³/3!    

We only take up to third derivative as, after that, all terms will be zero in this case.

Now, taking c = - 2, we get:

=> f(x) =  f(-2)  + f'(-2)(x +2)  + f''(-2)(x+2)² / 2! + f'''(-2)(x +2)³/3!      

f(-2) = 3(-2)³ - 2(-2)²+ (-2) + 4  = - 30

f"(-2) = 9(-2)² - 4(-2) + 1    = 45

f''(-2)  = 18(-2)  - 4  = - 40

f'''(-2)   = 18

=> f(x) =  -30  + 45(x +2) - 40(x+2)² / 2! + 18(x +2)³/3!  

=>  f(x) =  -30  + 45(x +2) - 20(x+2)² +  3(x +2)³

=> f(x) =  3(x +2)³  - 20(x+2)²  + 45(x +2) - 30

Therefore, the expansion of 3x³ - 2x²+ x - 4 = 3(x +2)³  - 20(x+2)²  + 45(x +2) - 30

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