Also, this one I have an answer but not rly sure so um help

Mathematics

1 answer:

mestny [16]2 years ago

7 0

Answer:

h(x) = - 2x² + 3

Step-by-step explanation:

Given f(x) then f(x) + c is a vertical translation of f(x)

• If c > 0 then a shift up of c units

• If c < 0 then a shift down of c units

Here the shift is 7 units up , then

h(x) = f(x) + 7 = - 2x² - 4 + 7 = - 2x² + 3

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Mrs. Computer is 3 times older than her daughter, Mousy. The sum of their ages is 52. How old is Mousy?

grandymaker [24]

Mousy=m
Mrs. Computer=3m
m+3m=52
4m=52
m=52/4
m=13
mousy=13, Mrs. Computer=3*13=39

8 0

3 years ago

I need help with part b. I feel like there’s a catch, I want to do the first derivative test, however, I feel like there is a be

Sladkaya [172]

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:

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

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