# 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?

Jul 16, 2018

$f ' \left(4.3\right) \approx - 3 + 10 \left(4.3 - 4\right) - 6 {\left(4.3 - 4\right)}^{2} = - 0.54$

#### Explanation:

First, find the derivative of $f \left(x\right)$

$f ' \left(x\right) = - 3 + 10 \left(x - 4\right) - 6 {\left(x - 4\right)}^{2} + 24 {\left(x - 4\right)}^{3}$

Next, find the second derivative of $f \left(x\right)$

$f ' ' \left(x\right) = 10 - 12 \left(x - 4\right) + 74 {\left(x - 4\right)}^{2}$

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 $f \left(4\right)$, $f ' \left(4\right)$, and $f ' ' \left(4\right)$

$f \left(4\right) = 7 - 3 \left(0\right) + 5 {\left(0\right)}^{2} - 2 {\left(0\right)}^{3} + 6 {\left(0\right)}^{4} = 7$

$f ' \left(4\right) = - 3 + 10 \left(0\right) - 6 {\left(0\right)}^{2} + 24 {\left(0\right)}^{3} = - 3$

$f ' ' \left(4\right) = 10 - 12 \left(0\right) + 74 {\left(0\right)}^{2} = 10$

Plugging these terms in to the above Taylor series gives

$7 - 3 {\left(x - 4\right)}^{1} + 5 {\left(x - 4\right)}^{2}$

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

$f \left(4.3\right) \approx 7 - 3 \left(4.3 - 4\right) + 5 {\left(4.3 - 4\right)}^{2} = 6.55$

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

$f ' \left(x\right) = g \left(x\right) = - 3 + 10 \left(x - 4\right) - 6 {\left(x - 4\right)}^{2} + 24 {\left(x - 4\right)}^{3}$

$g ' \left(x\right) = 10 - 12 \left(x - 4\right) + 72 {\left(x - 4\right)}^{2}$

$g ' ' \left(x\right) = - 12 + 144 \left(x - 4\right)$

At $x = 4$, these give

$g \left(4\right) = - 3$

$g ' \left(4\right) = 10$

$g ' ' \left(4\right) = - 12$

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

$= - 3 + 10 \left(x - 4\right) - 6 {\left(x - 4\right)}^{2}$

$g \left(4.3\right) = f ' \left(4.3\right) \approx - 3 + 10 \left(4.3 - 4\right) - 6 {\left(4.3 - 4\right)}^{2} = - 0.54$