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3 years ago

Find the derivative dy / dx for y = x^3 - 5x / x^2 - 1

1 answer:

soldier1979 [14.2K]3 years ago

8 0

$Use\\\\\left(\dfrac{h(x)}{g(x)}\right)'=\dfrac{h'(x)g(x)-h(x)g'(x)}{[g(x)]^2}\\\\\text{We have}\\\\h(x)=x^3-5x\to h'(x)=3x^2-5\\\\g(x)=x^2-1\to g'(x)=2x\\\\f(x)=\dfrac{x^3-5x}{x^2-1}\to f'(x)=\left(\dfrac{x^3-5x}{x^2-1}\right)'\\\\=\dfrac{(x^3-5x)'(x^2-1)-(x^3-5x)(x^2-1)'}{(x^2-1)^2}\\\\=\dfrac{(3x^2-5)(x^2-1)-(x^3-5x)(2x)}{(3x^2-5)^2}\\\\=\dfrac{3x^4-3x^2-5x^2+5-2x^4+10x^2}{(x^2-1)^2}\\\\=\dfrac{x^4+2x^2+5}{(x^2-1)^2}\\\\Answer:\ \left(\dfrac{x^3-5x}{x^2-1}\right)'=\dfrac{x^4+2x^2+5}{(x^2-1)^2}$

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3 years ago

1. Which point on the axis satisfies the inequality y

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Answer:

1) Point $(1,0)$ -----> see the attached figure N $1$

2) The value of x is $4$

3) I quadrant

4) $(1,1)$

5) $y>-5x+3$

Step-by-step explanation:

Part 1)

we know that

If the point satisfy the inequality

then

the point must be included in the shaded area

The point $(1,0)$ is included in the shaded area

Part 2)

we have

$x-2y\geq 4$

see the attached figure N $2$

we know that

The value for x on the boundary line and the x axis is equal to the x-intercept of the line $x-2y= 4$

For $y=0$

Find the value of x

$x-2(0)= 4$

$x=4$

The solution is $x=4$

Part 3)

we have

$x\geq 0$ -----> inequality A

The solution of the inequality A is in the first and fourth quadrant

$y\geq 0$ -----> inequality B

The solution of the inequality B is in the first and second quadrant

the solution of the inequality A and the inequality B is the first quadrant

Part 4) Which ordered pair is a solution of the inequality?

we have

$y\geq 4x-5$

we know that

If a ordered pair is a solution of the inequality

then

the ordered pair must be satisfy the inequality

we're going to verify all the cases

case A) point $(3,4)$

Substitute the value of x and y in the inequality

$x=3,y=4$

$4\geq 4(3)-5$

$4\geq 7$ ------> is not true

therefore

the point $(3,4)$ is not a solution of the inequality

case B) point $(2,1)$

Substitute the value of x and y in the inequality

$x=2,y=1$

$1\geq 4(2)-5$

$1\geq 3$ ------> is not true

therefore

the point $(2,1)$ is not a solution of the inequality

case C) point $(3,0)$

Substitute the value of x and y in the inequality

$x=3,y=0$

$0\geq 4(3)-5$

$0\geq 7$ ------> is not true

therefore

the point $(3,0)$ is not a solution of the inequality

case D) point $(1,1)$

Substitute the value of x and y in the inequality

$x=1,y=1$

$1\geq 4(1)-5$

$1\geq -1$ ------> is true

therefore

the point $(1,1)$ is a solution of the inequality

Part 5) Write an inequality to match the graph

we know that

The equation of the line has a negative slope

The y-intercept is the point $(3,0)$

The x-intercept is a positive number

The solution is the shaded area above the dashed line

the equation of the line is $y=-5x+3$

The inequality is $y>-5x+3$

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