Yes, you can do the derivative of the fifth degree Taylor polynomial, but notice that its derivative evaluated at x =-1 will give zero for all its terms except for the one of first order, so the calculation becomes simple:

$P_5(x)=g(-1)+g'(-1)\,(x+1)+g"(-1)\, \frac{(x+1)^2}{2!} +g^{(3)}(-1)\, \frac{(x+1)^3}{3!} + g^{(4)}(-1)\, \frac{(x+1)^4}{4!} +g^{(5)}(-1)\, \frac{(x+1)^5}{5!}$

and when you do its derivative:

1) the constant term renders zero,

2) the following term (term of order 1, the linear term) renders: $g'(-1)\,(1)$ since the derivative of (x+1) is one,

3) all other terms will keep at least one factor (x+1) in their derivative, and this evaluated at x = -1 will render zero

Therefore, the only term that would give you something different from zero once evaluated at x = -1 is the derivative of that linear term. and that only non-zero term is: $g'(-1)= 7$ as per the information given. Therefore, the function has derivative larger than zero, then it is increasing in the vicinity of x = -1

6 0

3 years ago

25 POINTS!

ludmilkaskok [199]

I would say it is <span>D- (-4,-5), for counterclockwise you turn towards the right. At 180 degrees it would be (4, -5) if you turn again you would get (-4,-5) hope this helps</span>

3 0

3 years ago

Geometric proofs. I just don’t understand it and my teacher isn’t helping.

pshichka [43]

The midpoint of a line can be represented by the point that is in the very center of the line. A line segment such as AT also represents half of the line. The symbol of the tilde with the equal sign underneath represents congruence meaning the two segments are the same. Therefore each equation shows the same true statement in a different form

5 0

2 years ago