Suppose T_4(x) = 7-3(x-2)+7(x-2)^2-6(x-2)^2+8(x-2)^4 is the degree 4 Taylor polynomial centered at x=2 for some function f, how do you estimate the value of f'(1.9)?

Aug 25, 2017

Depending on the final details in your approximation method, you could either say $f ' \left(1.9\right) \approx - 4.612$ (more accurate) or $f ' \left(1.9\right) \approx - 4.4$ (less accurate).

Explanation:

We have

$f \left(x\right) \approx {T}_{4} \left(x\right) = 7 - 3 \left(x - 2\right) + 7 {\left(x - 2\right)}^{2} - 6 {\left(x - 3\right)}^{3} + 8 {\left(x - 2\right)}^{4}$ for $x \approx 2$.

Therefore,

$f ' \left(x\right) \approx {T}_{4} ' \left(x\right) = - 3 + 14 \left(x - 2\right) - 18 {\left(x - 2\right)}^{2} + 32 {\left(x - 2\right)}^{3}$ for $x \approx 2$.

To get the more accurate approximation, plug $x = 1.9$ into this last equation:

$f ' \left(1.9\right) \approx {T}_{4} ' \left(1.9\right) = - 3 + 14 \cdot \left(- .1\right) - 18 \cdot {\left(- .1\right)}^{2} + 32 \cdot {\left(- .1\right)}^{3} \approx - 4.612$.

To get a less accurate approximation, we can get a linear approximation to ${T}_{4} ' \left(x\right)$ itself as ${T}_{4} ' \left(x\right) \approx - 3 + 14 \left(x - 2\right)$. Then $f ' \left(1.9\right) \approx {T}_{4} ' \left(1.9\right) \approx - 3 + 14 \cdot \left(- .1\right) = - 4.4$.