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3 years ago

H(x) = x^2 - 1 Over which interval does h have a negative average rate of change?

Mathematics

1 answer:

vredina [299]3 years ago

6 0

The Curve has negative rate of change for negative values of x

I.e in the interval - infinity to 0

Step-by-step explanation:

Step 1:

The change in the rate for the given function is obtained calculating the first derivative.

So dy/dx = dH(x)/dx = 2*x

Step 2:

This rate of change will be negative when the dy/ dx is less than 0

So dy/dx = 2x < 0 => x< 0

So for values of x less than 0 the function has negative values for the rate of change.

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2 years ago

Solve the equation<br> x-5=36-7(x+7)

lana66690 [7]

Answer: x = -1

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3 years ago

What are the solutions to the following system?

bija089 [108]

The solutions are $(\sqrt{2},-1) \text { and }(-\sqrt{2},-1)$ .

Solution:

Given equation:

$-2 x^{2}+y=-5$

Add $2 x^{2}$ on both sides.

$-2 x^{2}+y+2 x^{2}=-5+2 x^{2}$

$y=-5+2 x^{2}$ -------- (1)

$y=-3 x^{2}+5$ (given) -------- (2)

Equate (1) and (2).

$-5+2 x^{2}=-3 x^{2}+5$

Add $3 x^{2}$ on both sides.

$-5+2 x^{2}+3 x^{2}=-3 x^{2}+5 +3 x^{2}$

$-5+5x^{2}=5$

Add 5 on both sides.

$-5+5x^{2}+5=5+5$

$5x^{2}=10$

Divide by 5 on both sides, we get

$x^{2}=2$

Taking square root on both sides, we get

$x=\sqrt{2}, x=-\sqrt{2}$

Substitute $x=\sqrt{2}$ in (1).

$y=-5+2(\sqrt{2})^2$

$y=-5+4$

$y=-1$

Substitute $x=-\sqrt{2}$ in (1).

$y=-5+2(-\sqrt{2})^2$

$y=-5+4$

$y=-1$

Therefore the solutions are $(\sqrt{2},-1) \text { and }(-\sqrt{2},-1)$ .

Option C is the correct answer.

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