How do you do this question?

Step-by-step explanation:

f(x) = sin(x), a = π/6, n = 4

Find the derivatives.

f⁽⁰⁾(π/6) = sin(π/6) = ½

f⁽¹⁾(π/6) = cos(π/6) = ½√3

f⁽²⁾(π/6) = -sin(π/6) = -½

f⁽³⁾(π/6) = -cos(π/6) = -½√3

f⁽⁴⁾(π/6) = sin(π/6) = ½

T₄(x) = ½ (x − π/6)⁰ / 0! + ½√3 (x − π/6)¹ / 1! − ½ (x − π/6)² / 2! − ½√3 (x − π/6)³ / 3! + ½ (x − π/6)⁴ / 4!

T₄(x) = ½ + ½√3 (x − π/6) − ¼ (x − π/6)² − ¹/₁₂√3 (x − π/6)³ + ¹/₄₈ (x − π/6)⁴

Find the fifth derivative.

f⁽⁵⁾(x) = cos(x)

So the next term of the series would be:

cos(z) (x − π/6)⁵ / 5!

For 0 ≤ z ≤ π/3,│f⁽⁵⁾(x)│is a maximum at z = 0.  Therefore:

│R₄(x)│≤ │cos(0) (0 − π/6)⁵ / 5!│

│R₄(x)│≤ 0.000328

│R₄(x)│is always positive, so we can ignore the bottom two graphs.  The top right graph has a y-intercept of approximately 0.0003, so that is the correct graph.

option a.

step-by-step explanation:

if you look closely at the graph, we can notice that the function is not defined at two different points.

if you look at functions b and c, you will notice that both functions are not defined at just one point. the function b is not defined at x= -5/2 and function c is not defined at x=-5. both options are discarded, given that our function should not be defined at two fiferent points.

now, if you look closely at the graph you will notice that the function sketched is not defined at x=5 and x=-2.

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