How do you do this question?

Step-by-step explanation:

f(x) = ln(1+2x), a = 5, n = 3

Find the derivatives.

f⁽⁰⁾(5) = ln(1+2(5)) = ln(11)

f⁽¹⁾(5) = 2 / (1+2(5)) = 2/11

f⁽²⁾(5) = -4 / (1 + 2(5))² = -4/121

f⁽³⁾(5) = 16 / (1 + 2(5))³ = 16/1331

T₃(x) = ln(11) (x − 5)⁰ / 0! + 2/11 (x − 5)¹ / 1! − 4/121 (x − 5)² / 2! + 16/1331 (x − 5)³ / 3!

T₃(x) = ln(11) + 2/11 (x − 5) − 2/121 (x − 5)² − 8/3993 (x − 5)³

Find the fourth derivative.

f⁽⁴⁾(x) = -96 / (1 + 2x)⁴

So the next term of the series would be:

(-96 / (1 + 2x)⁴) (x − 5)⁴ / 4!

For 4.6 ≤ x ≤ 5.4,│f⁽⁴⁾(x)│is a maximum at x = 4.6.  Therefore:

│R₃(x)│≤ │(-96 / (1 + 2(4.6))⁴) (4.6 − 5)⁴ / 4!│

│R₃(x)│≤ 0.000009

│R₃(x)│is always positive, so we can ignore the right two graphs.  Only the top left graph is less than 9×10⁻⁶ on the interval.

the answer is ! ?