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2 years ago

For the given polynomial P(x) and c=1/2 use the remainder theorem to find P(c).

Mathematics

1 answer:

Aleonysh [2.5K]2 years ago

8 0

Answer:

-2.90625

Step-by-step explanation:

Given the polynomial function

P(x) = x^5-x^4+x^3-3

If c = 1/2

At x = c

P(1/2) = (1/2)^5-(1/2)^4+(1/2)^3-3

P(1/2) = 1/32 - 1/16 + 1/8 - 3

P(1/2) = 1-2+4-96/32

P(1/2) = -93/32

P(1/2) = -2.90625

Hence P(c) is -2.90625

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3 years ago

Consider the following. f(x) = x5 − x3 + 6, −1 ≤ x ≤ 1 (a) Use a graph to find the absolute maximum and minimum values of the fu

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ANSWER

See below

EXPLANATION

Part a)

The given function is

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From the graph, we can observe that, the absolute maximum occurs at (-0.7746,6.1859) and the absolute minimum occurs at (0.7746,5.8141).

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We plug this values into the original function to obtain the y-values of the turning points

$( - \frac{ \sqrt{15} }{5} , \frac{1}{125} ( 6 \sqrt{15} +750)) \:and \: (0, - 6) \: and\: ( \frac{ \sqrt{15} }{5} , \frac{1}{125} ( - 6 \sqrt{15} +750))$

We now use the second derivative test to determine the absolute maximum minimum on the interval [-1,1]

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Hence

$( - \frac{ \sqrt{15} }{5} , \frac{1}{125} ( 6 \sqrt{15} + 750))$

is a maximum point.

$f''( \frac{ \sqrt{15} }{5} ) \: > \: 0$

Hence

$( \frac{ \sqrt{15} }{5} , \frac{1}{125} (- 6 \sqrt{15} + 750))$

is a minimum point.

$f''(0) \: =\: 0$

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For the given polynomial​ P(x) and c=1/2 use the remainder theorem to find​ P(c).

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