3. Which of the following is a polynomial with roots 4,6, and - 7

Mathematics

1 answer:

Vladimir79 [104]3 years ago

5 0

Answer: P(x)= x³ - 3 x² - 46 x +168

Step-by-step explanation:

(4,6, -7)

hence the polynomial will be equal to

P(x) = (x-4) (x-6) (x+7)

P(x) = (x-4) (x²+7 x -6 x -42)

P(x) = (x-4) (x²+x -42)

P(x) = x³ + x²-42 x -4 x² -4 x +168

P(x) = x³ - 3 x² - 46 x +168

hence, the required polynomial is P(x) = x³ - 3 x² - 46 x +168

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Use the definition of continuity and the properties of limit to show that the function f(x)=x sqrtx/ (x-6)^2 is continuous at x=

jasenka [17]

Answer:

The function $\\ f(x) = \frac{x*\sqrt{x}}{(x-6)^{2}}$ is continuous at x = 36.

Step-by-step explanation:

We need to follow the following steps:

The function is:

$\\ f(x) = \frac{x*\sqrt{x}}{(x-6)^{2}}$

The function is continuous at point x=36 if:

The function $\\ f(x)$ exists at x=36.
The limit on both sides of 36 exists.
The value of the function at x=36 is the same as the value of the limit of the function at x = 36.

Therefore:

The value of the function at x = 36 is:

$\\ f(36) = \frac{36*\sqrt{36}}{(36-6)^{2}}$

$\\ f(36) = \frac{36*6}{900} = \frac{6}{25}$

The limit of the $\\ f(x)$ is the same at both sides of x=36, that is, the evaluation of the limit for values coming below x = 36, or 33, 34, 35.5, 35.9, 35.99999 is the same that the limit for values coming above x = 36, or 38, 37, 36.5, 36.1, 36.01, 36.001, 36.0001, etc.

For this case:

$\\ lim_{x \to 36} f(x) = \frac{x*\sqrt{x}}{(x-6)^{2}}$