Question

How do you find the second degree taylor polynomial for f' about 4 and use it to approximate f'(4.3) given `f(x) = 7 - 3(x-4) + 5(x-4)^2 - 2(x-4)^3 + 6(x-4)^4`?

Answer 1

`f'(4.3)~~-3+10(4.3-4) -6(4.3-4)^2=-0.54`

Explanation:

First, find the derivative of `f(x)`

`f'(x) = -3+10(x-4)-6(x-4)^2+24(x-4)^3`

Next, find the second derivative of `f(x)`

`f''(x)=10-12(x-4)+74(x-4)^2`

The formula for finding the Taylor series expansion of a function is

`f(a) + (f'(a))/(1!)(x-a)^1 + (f''(a))/(2!)(x-a)^2 + (f^(3)(a))/(3!)(x-a)^3+...`

Plugging in up to the second derivative term is all that is necessary because we only want the 2nd degree Taylor:

`f(4)+(f'(4))/(1!)(x-4)^1 + (f''(4))/(2!)(x-4)^2`

Lets get each of the terms `f(4)`, `f'(4)`, and `f''(4)`

`f(4) = 7-3(0)+5(0)^2-2(0)^3+6(0)^4=7`

`f'(4) = -3+10(0)-6(0)^2+24(0)^3 = -3`

`f''(4) = 10-12(0)+74(0)^2 = 10`

Plugging these terms in to the above Taylor series gives

`7-3(x-4)^1 + 5(x-4)^2`

If you had been asked to find the approximation of `f(x)` at `x=4.3`, you could now just plug in `x=4.3` into this, giving the approximation:

`f(4.3) ~~ 7-3(4.3-4)+5(4.3-4)^2= 6.55`

But Nay! You were asked to approximate `f'(x)` at `x=4.3`, not `f(4.3)`. At least that's the question above. So, starting with

`f'(x) = g(x) = -3+10(x-4)-6(x-4)^2+24(x-4)^3`

`g'(x) = 10-12(x-4)+72(x-4)^2`

`g''(x) = -12+144(x-4)`

At `x=4`, these give

`g(4) = -3`

`g'(4) = 10`

`g''(4) = -12`

Plugging these values in, gives

`g(4)+(g'(4))/(1!)(x-4)^1 + (g''(4))/(2!)(x-4)^2`

`=-3+(10)/(1!)(x-4)^1 + (-12)/(2!)(x-4)^2`

`=-3+10(x-4) -6(x-4)^2`

So, the final answer is

`g(4.3)=f'(4.3)~~-3+10(4.3-4) -6(4.3-4)^2=-0.54`