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3 years ago

Differentiate. Y=5x^2 +6x

1 answer:

5 0

The answer to this question is y' = 10x^2 + 6. You have to use the power rule to bring the exponent down as a coefficient and then reduce the exponent by 1. Do it for all the terms using the sum rule for derivatives (derivative of a sum of the sum of its derivatives).

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3 years ago

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cestrela7 [59]

Answer: 600

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4 years ago

9. y=-2(x-3)

Artyom0805 [142]

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4 years ago

Read 2 more answers

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

masya89 [10]

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:

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

8 0

3 years ago