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erastovalidia [21]

3 years ago

10

Write a polynomial function with the given zeros. multiply the terms (write in standard form).

1 answer:

sukhopar [10]3 years ago

3 0

Answer:

<h2>1)</h2><h2> $x^{3} - 3timesx -2 = 0$ </h2><h2>2)</h2><h2> $x^{3} - x^{2} - 6timesx$ </h2>

Step-by-step explanation:

1)

A polynomial function of x can be represented as

$f(x) = (x + 1)(x + 1)(x - 2)$ , if we put x = -1 / -1 / 2, then we will get x = 0.

Now,

$f(x) = (x + 1)(x + 1)(x - 2) = (x^{2} + 2timesx + 1)(x - 2) = x^{3} + 2timesx^{2} + x - 2timesx^{2} - 4timesx -2 = x^{3} - 3timesx -2.$

2)

Similarly, this function must have to be valued 0 for x = -2, 0, 3

Let g(x) be the function.

Hence,

$g(x) = (x + 2)(x - 0)(x - 3) = (x^{2} + 2timesx )(x - 3) = x^{3} + 2timesx^{2} - 3timesx^{2} -6timesx = x^{3} - x^{2} - 6timesx$

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1) increasing on (-∞,-1] ∪ [1,∞), decreasing on [-1,0) ∪ (0,1]

$x = -1$ is local maximum, $x = 1$ is local minimum

2) increasing on [1,∞), decreasing on (-∞,0) ∪ (0,1]

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3) increasing on (-∞,0] ∪ [8,∞), decreasing on [0,4) ∪ (4,8]

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4) increasing on [2,∞), decreasing on (-∞,2]

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Step-by-step explanation:

To find minima and maxima the of the function, we must take the derivative and equalize it to zero to find the roots.

1) $f(x) = 6x + 6/x$

$f\prime(x) = 6 - 6/x^2 = 0$ and $x \neq 0$

So, the roots are $x = -1$ and $x = 1$

The function is increasing on the interval (-∞,-1] ∪ [1,∞)

The function is decreasing on the interval [-1,0) ∪ (0,1]

$x = -1$ is local maximum, $x = 1$ is local minimum.

2) $f(x)=6-4/x+2/x^2$

$f\prime(x)=4/x^2-4/x^3=0$ and $x \neq 0$

So the root is $x = 1$

The function is increasing on the interval [1,∞)

The function is decreasing on the interval (-∞,0) ∪ (0,1]

$x = 1$ is absolute minimum.

3) $f(x) = 8x^2/(x-4)$

$f\prime(x) = (8x^2-64x)/(x-4)^2=0$ and $x \neq 4$

So the roots are $x = 0$ and $x = 8$

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The function is decreasing on the interval [0,4) ∪ (4,8]

$x = 0$ is local maximum, $x = 8$ is local minimum.

4) $f(x)=6(x-2)^{2/3} +4=0$

$f\prime(x) = 4/(x-2)^{1/3}$ has no solution and $x = 2$ is crtitical point.

The function is increasing on the interval [2,∞)

The function is decreasing on the interval (-∞,2]

$x = 2$ is absolute minimum.

5) $f(x)=8\sqrt x - 6x$ for $x>0$

$f\prime(x) = (4/\sqrt x)-6 = 0$

So the root is $x = 4/9$

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The function is decreasing on the interval [4/9,∞)

$x = 0$ is local minimum, $x = 4/9$ is absolute maximum.

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