All categories

3 years ago

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

You might be interested in

Find the equation of the line specified. The line passes through the points ( 5, 2) and ( 6, 4) a. y = 2x - 8 c. y = 2x + 12 b.

o-na [289]

( 5, 2) and ( 6, 4)
slope m = (4-2)/(6-5) = 2

y = mx + b
b = y - mx
b = 2 - 2(5)
b = 2 - 10
b = -8

so now you have slope m = 2 and y intercept b = -8

equation
y = 2x - 8

answer
a. y = 2x - 8

4 0

3 years ago

Read 2 more answers

Find T5(x) : Taylor polynomial of degree 5 of the function f(x)=cos(x) at a=0 . (You need to enter function.) T5(x)= Find all va

Burka [1]

Answer:

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

The polynomial is an approximation with an error less than or equals to 0.002652 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

f(x)=cos(x)

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

[-1.113826815, 1.113826815]

8 0

3 years ago

Find the equation for the horizontal line that goes through point M(a, b).

elena-14-01-66 [18.8K]

Horizontal means zero slope, or a zero coefficient on x.

Answer: y = b

6 0

3 years ago

Read 2 more answers

Carlos takes classes at both Westside Community College and Pinewood Community College. At Westside, class fees are $ 98 per cre

Sholpan [36]

Answer:

x + y = 13

98x + 115y = z

Step-by-step explanation:

Let x represent the time spent taking class in Westside community college

Let y represent the time spent taking class in Pinewood community college

Let z represent the combined total amount paid for the class fees.

The total combined time for the two school is given as:

x + y = 13.

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

Westside: $98 * x = 98x$

Pinewood: $115 * y = 115y$

The combined total amount represented as z is given as:

z = 98x + 115y

7 0

3 years ago

Part A: Factor x2y2 + 6xy2 + 8y2. Show your work. (4 points) Part B: Factor x2 + 8x + 16. Show your work. (3 points) Part C: Fac

skad [1K]

Answer:

Part A : y²(x + 2)(x + 4)

Part B: (x + 4) (x + 4)

Part C: (x + 4) (x - 4)

Step-by-step explanation:

Part A: Factor x²y²+ 6xy²+ 8y²

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)

6 0

3 years ago