What are the zeros of the function?

Mathematics

1 answer:

Yakvenalex [24]2 years ago

7 0

The real zeros are -8 and -9

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Find the Taylor series for f(x) centered at the given value of a. [Assume that f has a power series expansion. Do not show that

kherson [118]

Answer:

The answer is

$f(x) = {\displaystyle 8 + \frac{-28}{1}(x+2)+\frac{-36}{2!}(x+2)^2 + \frac{-18}{3!}(x+2)^2 }$

Step-by-step explanation:

Remember that Taylor says that

$f(x) = {\displaystyle \sum\limits_{k=0}^{\infty} \frac{f^{(k)}(a) }{k!}(x-a)^k }$

For this case

$f^{(0)} (-2) = 8(-2)-3(-2)^3 = 8\\f^{(1)} (-2) = 8-3(3)(-2)^2 = -28\\f^{(2)} (-2) = -3(3)2(-2) = -36\\f^{(2)} (-2) = -3(3)2 = -18$

$f(x) = {\displaystyle 8 + \frac{-28}{1}(x+2)+\frac{-36}{2!}(x+2)^2 + \frac{-18}{3!}(x+2)^2 }$

5 0

3 years ago

If sin theta = -1/2 and lies in the third quadrant, what is the cos?

never [62]

Sin angle that equals the 1/2 value is the 30 degree on the 1st Quadrant.

The angle of 30 degree on the 3rd Quadrant is equivalent to (formula and circle in the 1st photo) 180 + 30 = 210°.

So theta = 210° (3rd Quadrant) = 30° (1st Quadrant).

$cos \: 30\degree = \sqrt{3} \div 2$

Now just transfer the cos 30° to the 3rd Quadrant and the signal will be negative (as shown in the 2nd photo attached).

$cos \: 210 \degree = - \sqrt{3} \div 2$