M= -8 - -5/ 3 - 15= -8+5/3-15= -3/-12 = 3/12 = 1/4
The slope is 1/4.
The slope formula is m=y2-y1/x2-x1. I plugged in -8=y2, -5=y1, 3=x2, and 15=x1. Then I solved it.
Answer:
The fifth degree Taylor polynomial of g(x) is increasing around x=-1
Step-by-step explanation:
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!}](https://tex.z-dn.net/?f=P_5%28x%29%3Dg%28-1%29%2Bg%27%28-1%29%5C%2C%28x%2B1%29%2Bg%22%28-1%29%5C%2C%20%5Cfrac%7B%28x%2B1%29%5E2%7D%7B2%21%7D%20%2Bg%5E%7B%283%29%7D%28-1%29%5C%2C%20%5Cfrac%7B%28x%2B1%29%5E3%7D%7B3%21%7D%20%2B%20g%5E%7B%284%29%7D%28-1%29%5C%2C%20%5Cfrac%7B%28x%2B1%29%5E4%7D%7B4%21%7D%20%2Bg%5E%7B%285%29%7D%28-1%29%5C%2C%20%5Cfrac%7B%28x%2B1%29%5E5%7D%7B5%21%7D)
and when you do its derivative:
1) the constant term renders zero,
2) the following term (term of order 1, the linear term) renders:
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:
as per the information given. Therefore, the function has derivative larger than zero, then it is increasing in the vicinity of x = -1
150% of 84 i believe lol .
It is a knowledge check
On aleks