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Oduvanchick [21]
3 years ago
14

If ,f(x)=3x-6/x-2 what is the average rate of change of f(x) over the interval [6, 8]?

Mathematics
2 answers:
sukhopar [10]3 years ago
4 0
I hope this helps you



f (x)= 3 (x-2)/(x-2)


f (x)= 3


f(6)= 3


f (8)= 3
enyata [817]3 years ago
3 0

Answer:

the average rate of change of f(x) over the interval [6, 8] is, 0

Step-by-step explanation:

Average rate of change (A(x)) of f(x) over interval [a, b] is given by:

A(x) = \frac{f(b)-f(a)}{b-a}       ....[1]

As per the statement:

Given the function:

f(x) = \frac{3x-6}{x-2}

At x = 6

f(6) = \frac{3(6)-6}{6-2}=\frac{18-6}{4}=\frac{12}{4} =3

⇒f(6) = 3

At x = 8

f(8) = \frac{3(8)-6}{8-2}=\frac{24-6}{6}=\frac{18}{6} =3

⇒f(8) = 3

We have to find the average rate of change of f(x) over the interval [6, 8]

Substitute the given values in [1] we have;

A(x) = \frac{f(8)-f(6)}{8-6}=\frac{3-3}{2}=\frac{0}{2} = 0

Therefore, the average rate of change of f(x) over the interval [6, 8] is, 0

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<span>( 5, 2) and ( 6, 4)
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Answer:

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The polynomial is an approximation with an error less than or equals to <em>0.002652</em> for x in the interval

[-1.113826815, 1.113826815]

Step-by-step explanation:

According to Taylor's theorem

\bf f(x)=f(0)+f'(0)x+f''(0)\displaystyle\frac{x^2}{2}+f^{(3)}(0)\displaystyle\frac{x^3}{3!}+f^{(4)}(0)\displaystyle\frac{x^4}{4!}+f^{(5)}(0)\displaystyle\frac{x^5}{5!}+R_6(x)

with

\bf R_6(x)=f^{(6)}(c)\displaystyle\frac{x^6}{6!}

for some c in the interval (-x, x)

In the particular case f

<em>f(x)=cos(x) </em>

<em> </em>

we have

\bf f'(x)=-sin(x)\\f''(x)=-cos(x)\\f^{(3)}(x)=sin(x)\\f^{(4)}(x)=cos(x)\\f^{(5)}(x)=-sin(x)\\f^{(6)}(x)=-cos(x)

therefore

\bf f'(x)=-sin(0)=0\\f''(0)=-cos(0)=-1\\f^{(3)}(0)=sin(0)=0\\f^{(4)}(0)=cos(0)=1\\f^{(5)}(0)=-sin(0)=0

and the polynomial approximation of T5(x) of cos(x) would be

\bf cos(x)\approx1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{4!}=\\\\=1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{24}

In order to find all the values of x for which this approximation is within 0.002652 of the right answer, we notice that

\bf R_6(x)=-cos(c)\displaystyle\frac{x^6}{6!}

for some c in (-x,x). So

\bf |R_6(x)|\leq|\displaystyle\frac{x^6}{6!}|=\displaystyle\frac{|x|^6}{6!}

and we must find the values of x for which

\bf \displaystyle\frac{|x|^6}{6!}\leq0.002652

Working this inequality out, we find

\bf \displaystyle\frac{|x|^6}{6!}\leq0.002652\Rightarrow |x|^6\leq1.90944\Rightarrow\\\\\Rightarrow |x|\leq\sqrt[6]{1.90944}\Rightarrow |x|\leq1.113826815

Therefore the polynomial is an approximation with an error less than or equals to 0.002652 for x in the interval

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For the combined total amount; we multiply the price for each school with the time spent on the school and sum them together.

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The combined total amount represented as z is given as:

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Answer:

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Step-by-step explanation:

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x²y²+ 6xy²+ 8y²

y² is very common across the quadratic equation , hence

= y² (x² + 6x + 8)

= (y²) (x² + 6x + 8)

= (y²) (x² + 2x +4x + 8)

= (y²) (x² + 2x)+(4x + 8)

= (y²) (x(x + 2)+ 4(x + 2))

= y²(x+2)(x+4)

Part B: Factor x² + 8x + 16

x² + 8x + 16

= x² + 4x + 4x + 16

= (x² + 4x) + (4x + 16)

= x( x + 4) + 4(x + 4)

= (x + 4) (x + 4)

Part C: Factor x² − 16

= x² − 16

= x² + 0x − 16

= x² + 4x - 4x - 16

= (x² + 4x) - (4x - 16)

= x (x + 4) - 4(x + 4)

= (x + 4) (x - 4)

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