# Taylor Polynomials question

• stunner5000pt
You're right that you'll need to take derivatives and set them to zero, but you're working with the wrong function.In summary, the conversation discusses finding the Taylor polynomial and remainder for the function f(x) = 4 + 5x - 6x^2 + 11x^3 - 19x^4 + x^5, as well as finding the maximum value of the third derivative of the function on the interval |x| < 0.1. The group also discusses the process for finding a maximum value and the importance of considering endpoints when there are no local maxima within the interval.

#### stunner5000pt

Homework Statement
$$f(x) = 4 + 5x - 6x^2 + 11x^3 - 19x^4 + x^5$$
a. Find Taylor polynomial at x = 0, order 2
b. find the remainder
$$R_{2} (x) = f(x) - T_{2} (x)$$
c. Find the maximum values of $$f^{(3)} (x)$$ on the interval |x| < 0.1
Relevant Equations
Taylor polynomial formula
$$f(x) = 4 + 5x - 6x^2 + 11x^3 - 19x^4 + x^5$$

question a almost seems too easy as I end up 'removing' the x^4 and x^5 terms
a.
$$T_{2} (x) = 4 + 5x - 6x^2$$

b.
$$= R_{2} (x) = 11x^3 - 19x^4 + x^5$$

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?

stunner5000pt said:
Homework Statement:: $$f(x) = 4 + 5x - 6x^2 + 11x^3 - 19x^4 + x^5$$
a. Find Taylor polynomial at x = 0, order 2
b. find the remainder
$$R_{2} (x) = f(x) - T_{2} (x)$$
c. Find the maximum values of $$f^{(3)} (x)$$ on the interval |x| < 0.1
Relevant Equations:: Taylor polynomial formula

$$f(x) = 4 + 5x - 6x^2 + 11x^3 - 19x^4 + x^5$$

question a almost seems too easy as I end up 'removing' the x^4 and x^5 terms
a.
$$T_{2} (x) = 4 + 5x - 6x^2$$

b.
$$= R_{2} (x) = 11x^3 - 19x^4 + x^5$$
Yes, I think it is that easy.
stunner5000pt said:
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
That's not ##f^{(3)}(x)##.
stunner5000pt said:
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?
If there are is no local maximum within an interval, then the maximum value must be at an endpoint.

The answer to your question is yes. To find a maximum, you check the critical points and the endpoints.

PeroK said:
Yes, I think it is that easy.

That's not ##f^{(3)}(x)##.

If there are is no local maximum within an interval, then the maximum value must be at an endpoint.
Right, i see the issue. THe maximum of f'''(x) would be solved by solving f''(x) = 0, is that correct?

stunner5000pt said:
Right, i see the issue. THe maximum of f'''(x) would be solved by solving f''(x) = 0, is that correct?
No ##f^{(4)}(x) = 0##.

stunner5000pt said:
Right, i see the issue. THe maximum of f'''(x) would be solved by solving f''(x) = 0, is that correct?

No, you need to take one extra derivative, so you need to fourth derivative.

It might help to first compute ##f^{(3)}## and then start from scratch on that. It's a function, and you need to maximize it

• PeroK