All categories

3 years ago

Calculate the average rate of change for the function f(x) = x4 + 3x3 - 5x2 + 2x - 2, from x = -1 to x = 1

Mathematics

2 answers:

kolezko [41]3 years ago

5 0

In interval <a, b> the average rate of change the function you count from this formula:
$x=\frac{f(b)-f(a)}{b-a} \qquad b>a$
Here you've got a=-1 and b=1:
$f(a)=f(-1)=(-1)^4+3 \cdot (-1)^3 - 5 \cdot (-1)^2 +2 \cdot (-1)-2= \\ = 1-3-5-2-2=-11 \\ \\ f(b)=f(1)=1^4+3 \cdot 1^4 - 5 \cdot 1^2 + 2 \cdot 1 -2=1+3-5+2-2=-1 \\ \\ \hbox{So average rate of change this function is:} \\ \\ x= \frac{-1+11}{1+1}=\frac{10}{2}=5$

Georgia [21]3 years ago

3 0

Workings and answer in the attachments below.

You might be interested in

250 gramı 3.75 TL olan peynirin 1 kilogramı kaç liradır?

DENIUS [597]

Answer:

3.75•4=15 TL dir .......

6 0

3 years ago

WOULD U RATHER?? Please show your work Which one would u choose?

erik [133]

The last one because it has inches, and inches are bigger than centimeters.

8 0

4 years ago

I need help please :/

OleMash [197]

Part (a)

Plug in y = 0 and solve for x. Use the zero product property

y = x(x+3)(x-2)

0 = x(x+3)(x-2)

x(x+3)(x-2) = 0

x = 0 or x+3 = 0 or x-2 = 0 .... zero product property

x = 0 or x = -3 or x = 2

The three roots or zeros are x = 0 or x = -3 or x = 2

================================================

Part (b)

The roots of the graph are: x = -2, x = 1, x = 3. Each root is where the graph crosses or touches the horizontal x axis.

Note how x = 0 is found in part (a), but not found here. This is one example where the graphs don't match. Another would be x = -3 is in part (a), but not here.

So that's why the graph does not match with the function in part (a)

7 0

3 years ago

Factor the expression using GCF

julsineya [31]

7 + 14
7(1 + 2)

44 - 11
11( 4 - 1)

18 - 12
6(3 - 2)

70 + 95
5(14 + 19)

60 - 36
12(5 - 3)

100 - 80
20(5 - 4)

6 0

3 years ago

The following table shows the percent increase of donations made on behalf of a non-profit organization for the period of 1984 t

pashok25 [27]

Giving the table below which shows the percent increase of donations made on behalf of a non-profit organization for the period of 1984 to 2003.

Year: 1984 1989 1993 1997 2001 2003

Percent: 7.8 16.3 26.2 38.9 49.2 62.1

The scatter plot of the data is attached with the x-axis representing the number of years after 1980 and the y-axis representing the percent increase of donations made on behalf of a non-profit organization.

To find the equation for the line of regression where the x-axis representing the number of years after 1980 and the y-axis representing the percent increase of donations made on behalf of a non-profit organization.
$\begin{center}
\begin{tabular}{ c| c| c| c| }
 x & y & x^2 & xy \\ [1ex] 
 4 & 7.8 & 16 & 31.2 \\ 
 9 & 16.3 & 81 & 146.7 \\ 
13 & 26.2 & 169 & 340.6 \\ 
17 & 38.9 & 289 & 661.3 \\ 
21 & 49.2 & 441 & 1,033.2 \\ 
23 & 62.1 & 529 & 1,428.3 \\ [1ex]
\Sigma x=87 & \Sigma y=200.5 & \Sigma x^2=1,525 & \Sigma xy=3,641.3 
\end{tabular}
\end{center}$

Recall that the equation of the regression line is given by
$y=a+bx$
where
$a= \frac{(\Sigma y)(\Sigma x^2)-(\Sigma x)(\Sigma xy)}{n(\Sigma x^2)-(\Sigma x)^2} = \frac{200.5(1,525)-87(3,641.3)}{6(1,525)-(87)^2} \\ \\ = \frac{305,762.5-316793.1}{9,150-7,569} = \frac{-11,030.6}{1,581} =-6.977$
and
$b= \frac{n(\Sigma xy)-(\Sigma x)(\Sigma y)}{n(\Sigma x^2)-(\Sigma x)^2} = \frac{6(3,641.3)-(87)(200.5)}{6(1,525)-(87)^2} \\ \\ = \frac{21,847.8-17,443.5}{9,150-7,569} = \frac{4,404.3}{1,581} =2.7858$

Thus, the equation of the regresson line is given by
$y=-6.977+2.7858x$

The graph of the regression line is attached.

Using the equation, we can predict the percent donated in the year 2015. Recall that 2015 is 35 years after 1980. Thus x = 35.

The percent donated in the year 2015 is given by
$-6.977+2.7858(35)=-6.977+97.503=90.526$

Therefore, the percent donated in the year 2015 is predicted to be 90.5

7 0

3 years ago