When solving a quadratic, how can x equal 2 different things? What is x?

Mathematics

1 answer:

melisa1 [442]3 years ago

4 0

Answer:

X = unknown

The highest degree of x is 2 .

Step-by-step explanation:

First of all x is the unknown variable to be determined by solving quadratically

Quadratic equation was chooses because of the degree of x.

The highest degree of x is 2 .

That's the the reason why the value of x will always be double.

If solving linear equation that has degree one, the value of x is one

If solving polynomial, the value of x will be 3 above.

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Custom Office makes a line of executive desks. It is estimated that the total cost for making x units of their Senior Executive

Ivan

Answer:

(a) The average cost function is $\bar{C}(x)=95+\frac{230000}{x}$

(b) The marginal average cost function is $\bar{C}'(x)=-\frac{230000}{x^2}$

(c) The average cost approaches to 95 if the production level is very high.

Step-by-step explanation:

(a) Suppose $C(x)$ is a total cost function. Then the average cost function, denoted by $\bar{C}(x)$ , is

$\frac{C(x)}{x}$

We know that the total cost for making x units of their Senior Executive model is given by the function

$C(x) = 95x + 230000$

The average cost function is

$\bar{C}(x)=\frac{C(x)}{x}=\frac{95x + 230000}{x} \\\bar{C}(x)=95+\frac{230000}{x}$

(b) The derivative $\bar{C}'(x)$ of the average cost function, called the marginal average cost function, measures the rate of change of the average cost function with respect to the number of units produced.

The marginal average cost function is

$\bar{C}'(x)=\frac{d}{dx}\left(95+\frac{230000}{x}\right)\\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g\\\\\frac{d}{dx}\left(95\right)+\frac{d}{dx}\left(\frac{230000}{x}\right)\\\\\bar{C}'(x)=-\frac{230000}{x^2}$

(c) The average cost approaches to 95 if the production level is very high.

$\lim_{x \to \infty} (\bar{C}(x))=\lim_{x \to \infty} (95+\frac{230000}{x})\\\\\lim _{x\to a}\left[f\left(x\right)\pm g\left(x\right)\right]=\lim _{x\to a}f\left(x\right)\pm \lim _{x\to a}g\left(x\right)\\\\=\lim _{x\to \infty \:}\left(95\right)+\lim _{x\to \infty \:}\left(\frac{230000}{x}\right)\\\\\lim _{x\to a}c=c\\\lim _{x\to \infty \:}\left(95\right)=95\\\\\mathrm{Apply\:Infinity\:Property:}\:\lim _{x\to \infty }\left(\frac{c}{x^a}\right)=0\\\lim_{x \to \infty} (\frac{230000}{x} )=0$