How Do You Simplify and Analyze Taylor Polynomials for Higher Degree Functions?

[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?

 

[PeroK]

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.

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 $f^{(3)}(x)$.

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.

 

[vela]

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

 

[stunner5000pt]

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?

 

[PeroK]

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

No $f^{(4)}(x) = 0$.

 

[Office_Shredder]

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