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MissTica
3 years ago
11

Halla la tasa de variación de cada funcion en el intervalo [-4,3] e indica si es positiva , negativa o nula A) f(x)=x2-2x+4 B) f

(x)= -3x + 2 En rl ejercico A es x elevado al 2 o cuadrado
Mathematics
1 answer:
masya89 [10]3 years ago
8 0

Answer:

A) \hspace{3}Rate\hspace{3}of\hspace{3}change=-5\hspace{3}Negative\\\\B)\hspace{3}Rate\hspace{3}of\hspace{3}change=-21\hspace{3}Negative  

Step-by-step explanation:

Given a function f(x), we called the rate of change to the number that represents the increase or decrease that the function experiences when increasing the independent variable from one value "x_1" to another "x_2".

The rate of change of f(x) between x_1 and x_2 can be calculated as follows:

Rate\hspace{3}of\hspace{3}change=f(x_2)-f(x_1)

For:

f(x)=x^2-2x+4

Let's find f(x_1) and f(x_2), where:

[x_1,x_2]=[-4,3]

f(x_1)=f(-4)=(-4)^2-2(4)+4=16-8+4=12\\f(x_2)=f(3)=(3)^2-2(3)+4=9-6+4=7

So:

Rate\hspace{3}of\hspace{3}change =7-12=-5\hspace{3}Negative

And for:

f(x)-3x+2

Let's find f(x_1) and f(x_2), where:

[x_1,x_2]=[-4,3]

f(x_1)=f(-4)=-3(-4)+2=12+2=14\\f(x_2)=f(3)=-3(3)+2=-9+2=-7

So:

Rate\hspace{3}of\hspace{3}change =-7-14=-21\hspace{3}Negative

<em>Translation:</em>

Dada una función f(x), llamábamos tasa de variación al número que representa el aumento o disminución que experimenta la función al aumentar la variable independiente de un valor "x_1" a otro "x_2".

La tasa de variación de f(x) entre x_1 y x_2, puede ser calculada de la siguiente forma:

Tasa\hspace{3}de\hspace{3}variacion=f(x_2)-f(x_1)

Para:

f(x)=x^2-2x+4

Encontremos f(x_1) y f(x_2), donde:

[x_1,x_2]=[-4,3]

f(x_1)=f(-4)=-3(-4)+2=12+2=14\\f(x_2)=f(3)=-3(3)+2=-9+2=-7

Entonces:

Tasa\hspace{3}de\hspace{3}variacion =7-12=-5\hspace{3}Negativa

Y para:

f(x)-3x+2

Encontremos f(x_1) y f(x_2), donde:

[x_1,x_2]=[-4,3]

f(x_1)=f(-4)=-3(-4)+2=12+2=14\\f(x_2)=f(3)=-3(3)+2=-9+2=-7

Entonces:

Tasa\hspace{3}de\hspace{3}variacion=-7-14=-21\hspace{3}Negativa

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  1. C) x = plus/minus 11
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  1. Note how (11)^2 = (11)*(11) = 121 and also (-11)^2 = (-11)*(-11) = 121. The two negatives multiply to a positive. So that's why the solution is x = plus/minus 11. The plus minus breaks down into the two equations x = 11 or x = -11.
  2. There are no real solutions here because the left hand side can never be negative, no matter what real number you pick for x. As mentioned in problem 1, squaring -11 leads to a positive number 121. The same idea applies here as well.
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