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2 years ago

F(x)=x^3−9x

Mathematics

1 answer:

Damm [24]2 years ago

8 0

Option B

Over the [-4,-1], the function has positive rate of change

<em><u>Solution:</u></em>

<em><u>The average rate of change is given by formula:</u></em>

$Rate\ of\ change = \frac{f(b)-f(a)}{b-a}$

Given function is:

$f(x) = x^3-9x$

<h3><em><u>Option A</u></em></h3>

[-2,1]

Find f(-2) and f(1)

Substitute x = -2 in given function

$f(-2) = (-2)^3 -9(-2)\\\\f(-2) = -8 + 18 = 10$

Substitute x = 1 in given function

$f(1) = 1^3-9(1) = 1-9 = -8$

Find rate of change:

$Rate\ of\ change = \frac{f(-2)-f(1)}{-2-1}\\\\Rate\ of\ change = \frac{10-(-8)}{-3} = \frac{18}{-3} = -6$

Thus this has a negative rate of change

<h3><u>Option B</u></h3>

[-4,-1]

Find f(-4) and f(-1)

Substitute x = -4 in given function

$f(-4) = (-4)^3 -9 (-4) = -64 + 36 = -28$

Substitute x = -1 in given function

$f(-1) = (-1)^3 -9(-1)\\\\f(-1) = -1 + 9 = 8$

Find rate of change:

$Rate\ of\ change = \frac{8-(-28)}{-1-(-4)} = \frac{36}{3} = 12$

Thus this has a positive rate of change

<h3><u>Option C</u></h3>

[-1,2]

Substitute x = -1 in given function

$f(-1) = (-1)^3 -9(-1)\\\\f(-1) = -1 + 9 = 8$

Substitute x = 2 in given function

$f(2) = 2^3 -9(2) = 8-18 = -10$

Find rate of change:

$Rate\ of\ change = \frac{-10-8}{2-(-1)} = \frac{-18}{3} = -6$

Thus this has a negative rate of change

<h3><u>Option D</u></h3>

[-3,3]

Substitute x = -3 in given functiom

$f(-3) = (-3)^3 - 9(-3) = -27 + 27 = 0$

Substitute x = 3 in given function

$f(3) = 3^3 -9(3) = 27 - 27 = 0$

Find rate of change:

$Rate\ of\ change = \frac{0-0}{3-(-3)} = 0$

Thus rate of change is 0

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1.) Find the length of the arc of the graph x^4 = y^6 from x = 1 to x = 8.

xxTIMURxx [149]

First, rewrite the equation so that <em>y</em> is a function of <em>x</em> :

$x^4 = y^6 \implies \left(x^4\right)^{1/6} = \left(y^6\right)^{1/6} \implies x^{4/6} = y^{6/6} \implies y = x^{2/3}$

(If you were to plot the actual curve, you would have both $y=x^{2/3}$ and $y=-x^{2/3}$ , but one curve is a reflection of the other, so the arc length for 1 ≤ <em>x</em> ≤ 8 would be the same on both curves. It doesn't matter which "half-curve" you choose to work with.)

The arc length is then given by the definite integral,

$\displaystyle \int_1^8 \sqrt{1 + \left(\frac{\mathrm dy}{\mathrm dx}\right)^2}\,\mathrm dx$

We have

$y = x^{2/3} \implies \dfrac{\mathrm dy}{\mathrm dx} = \dfrac23x^{-1/3} \implies \left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2 = \dfrac49x^{-2/3}$

Then in the integral,

$\displaystyle \int_1^8 \sqrt{1 + \frac49x^{-2/3}}\,\mathrm dx = \int_1^8 \sqrt{\frac49x^{-2/3}}\sqrt{\frac94x^{2/3}+1}\,\mathrm dx = \int_1^8 \frac23x^{-1/3} \sqrt{\frac94x^{2/3}+1}\,\mathrm dx$

Substitute

$u = \dfrac94x^{2/3}+1 \text{ and } \mathrm du = \dfrac{18}{12}x^{-1/3}\,\mathrm dx = \dfrac32x^{-1/3}\,\mathrm dx$

This transforms the integral to

$\displaystyle \frac49 \int_{13/4}^{10} \sqrt{u}\,\mathrm du$

and computing it is trivial:

$\displaystyle \frac49 \int_{13/4}^{10} u^{1/2} \,\mathrm du = \frac49\cdot\frac23 u^{3/2}\bigg|_{13/4}^{10} = \frac8{27} \left(10^{3/2} - \left(\frac{13}4\right)^{3/2}\right)$