- #1

- 1,457

- 2

- Homework Statement
- [tex] f(x) = 4 + 5x - 6x^2 + 11x^3 - 19x^4 + x^5 [/tex]

a. Find Taylor polynomial at x = 0, order 2

b. find the remainder

[tex] R_{2} (x) = f(x) - T_{2} (x) [/tex]

c. Find the maximum values of [tex] f^{(3)} (x) [/tex] on the interval |x| < 0.1

- Relevant Equations
- Taylor polynomial formula

[tex] f(x) = 4 + 5x - 6x^2 + 11x^3 - 19x^4 + x^5 [/tex]

question a almost seems too easy as I end up 'removing' the x^4 and x^5 terms

a.

[tex] T_{2} (x) = 4 + 5x - 6x^2 [/tex]

b.

[tex] = R_{2} (x) = 11x^3 - 19x^4 + x^5 [/tex]

c.

i don't understand what i need to do here. To find the maximum value of a function, we differentiate and make that derivative = 0

so if we are to find the maximum of f'''(x) , does that mean that we simply make the answer from a = 0?

4 + 5x - 6x^2 = 0

This solves to

x= -1/2 and x = 4/3

But since neither of these values is in the given interval of |x| < 0.1, do we just evaluate T(2) (x) at x = -0.1 and x = 0.1 and determine the larger of the two?

question a almost seems too easy as I end up 'removing' the x^4 and x^5 terms

a.

[tex] T_{2} (x) = 4 + 5x - 6x^2 [/tex]

b.

[tex] = R_{2} (x) = 11x^3 - 19x^4 + x^5 [/tex]

c.

i don't understand what i need to do here. To find the maximum value of a function, we differentiate and make that derivative = 0

so if we are to find the maximum of f'''(x) , does that mean that we simply make the answer from a = 0?

4 + 5x - 6x^2 = 0

This solves to

x= -1/2 and x = 4/3

But since neither of these values is in the given interval of |x| < 0.1, do we just evaluate T(2) (x) at x = -0.1 and x = 0.1 and determine the larger of the two?