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2 years ago

6

Based on the graph given for f (x) construct the graph of f¹ (x). for f (x) find E, F and the formula of f (x) when it is known

that it is a linear function.

1 answer:

Lostsunrise [7]2 years ago

4 0

Answer:

To find an inverse function, reflect a graph of a function across the line y=x (and find the resulting equation)

To reflect a linear function in the line y=x, find points on f(x) and then swap their x and y coordinates.

Points on f(x): (0, -4) (2, 0) (5, 6)

Points reflected in line y=x: (-4, 0) (0, 2) (6, 5)

Plot points (-4, 0) (0, 2) (6, 5) and connect to form a straight line - this is the inverse of the function: $f^{-1}(x)$

To determine the equation of f(x):

Choose 2 points on f(x): (5, 6) and (0, -4)

Calculate the slope (gradient) by using:

$m=\dfrac{\triangle y}{\triangle x}=\dfrac{y_2-y_1}{x_2-x_1}=\frac{6--4}{5-0}=2$

Using the slope-intercept form: y = mx + b

(where m is the slope and b is in the y-intercept)

From inspection of the graph, we can see the line crosses the y-axis at -4,

⇒ f(x) = 2x - 4

As the line is actually a line segment (with endpoints (0, -4) and (5, 6), then

f(x) = 2x - 4, 0 ≤ x ≤ 5

To determine the equation of $f^{-1}(x)$ :

Rewrite f(x) as $y = 2x - 4$

Swap the x and y: $x = 2y - 4$

Rearrange to make y the subject: $y= \dfrac{1}{2}(x + 4)$

Replace y with $f^{-1}(x)$

So the equation of the inverse is: $f^{-1}(x)= \dfrac{1}{2}(x + 4)$

As the original function is a segment, then

$f^{-1}(x)= \dfrac{1}{2}(x + 4), \ \ \ -4\leq x\leq 6$

(shown in blue on the attached diagram)

** I can't see any points labelled E and F on the original function. If they are the endpoints of the line segment of f(x), then they are (0, -4) and (5, 6) **

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An art class is making a mural for the school that has a triangle drawn in the middle. The length of the bottom of the triangle

Arte-miy333 [17]

Answer:

7

Step-by-step explanation:

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3 years ago

Lenovo uses the zx-81 chip in some of its laptop computers. the prices for the chip during the last 12 months were as follows:

Stella [2.4K]

Given the table below of the prices for the Lenovo zx-81 chip during the last 12 months

$\begin{tabular}
{|c|c|c|c|}
Month&Price per Chip&Month&Price per Chip\\[1ex]
January&\$1.90&July&\$1.80\\
February&\$1.61&August&\$1.83\\
March&\$1.60&September&\$1.60\\
April&\$1.85&October&\$1.57\\
May&\$1.90&November&\$1.62\\
June&\$1.95&December&\$1.75
\end{tabular}$

The forcast for a period $F_{t+1}$ is given by the formular

$F_{t+1}=\alpha A_t+(1-\alpha)F_t$

where $A_t$ is the actual value for the preceding period and $F_t$ is the forcast for the preceding period.

Part 1A:
Given <span>α = 0.1 and the initial forecast for october of $1.83, the actual value for october is $1.57.

Thus, the forecast for period 11 is given by:

$F_{11}=\alpha A_{10}+(1-\alpha)F_{10} \\ \\ =0.1(1.57)+(1-0.1)(1.83) \\ \\ =0.157+0.9(1.83)=0.157+1.647 \\ \\ =1.804$

Therefore, the foreast for period 11 is $1.80

Part 1B:

</span>Given <span>α = 0.1 and the forecast for november of $1.80, the actual value for november is $1.62

Thus, the forecast for period 12 is given by:

$F_{12}=\alpha
 A_{11}+(1-\alpha)F_{11} \\ \\ =0.1(1.62)+(1-0.1)(1.80) \\ \\ 
=0.162+0.9(1.80)=0.162+1.62 \\ \\ =1.782$

Therefore, the foreast for period 12 is $1.78</span>

Part 2A:

Given <span>α = 0.3 and the initial forecast for october of $1.76, the actual value for October is $1.57.

Thus, the forecast for period 11 is given by:

$F_{11}=\alpha
 A_{10}+(1-\alpha)F_{10} \\ \\ =0.3(1.57)+(1-0.3)(1.76) \\ \\ 
=0.471+0.7(1.76)=0.471+1.232 \\ \\ =1.703$

Therefore, the foreast for period 11 is $1.70

</span>
<span><span>Part 2B:

</span>Given <span>α = 0.3 and the forecast for November of $1.70, the actual value for november is $1.62

Thus, the forecast for period 12 is given by:

$F_{12}=\alpha
 A_{11}+(1-\alpha)F_{11} \\ \\ =0.3(1.62)+(1-0.3)(1.70) \\ \\ 
=0.486+0.7(1.70)=0.486+1.19 \\ \\ =1.676$

Therefore, the foreast for period 12 is $1.68

</span></span>
<span>Part 3A:

Given <span>α = 0.5 and the initial forecast for october of $1.72, the actual value for October is $1.57.

Thus, the forecast for period 11 is given by:

$F_{11}=\alpha
 A_{10}+(1-\alpha)F_{10} \\ \\ =0.5(1.57)+(1-0.5)(1.72) \\ \\ 
=0.785+0.5(1.72)=0.785+0.86 \\ \\ =1.645$

Therefore, the forecast for period 11 is $1.65

</span>
<span><span>Part 3B:

</span>Given <span>α = 0.5 and the forecast for November of $1.65, the actual value for November is $1.62

Thus, the forecast for period 12 is given by:

$F_{12}=\alpha
 A_{11}+(1-\alpha)F_{11} \\ \\ =0.5(1.62)+(1-0.5)(1.65) \\ \\ 
=0.81+0.5(1.65)=0.81+0.825 \\ \\ =1.635$

Therefore, the forecast for period 12 is $1.64

Part 4:

The mean absolute deviation of a forecast is given by the summation of the absolute values of the actual values minus the forecasted values all divided by the number of items.

Thus, given that the actual values of october, november and december are: $1.57, $1.62, $1.75

using </span></span></span><span>α = 0.3, we obtained that the forcasted values of october, november and december are: $1.83, $1.80, $1.78

Thus, the mean absolute deviation is given by:

$\frac{|1.57-1.83|+|1.62-1.80|+|1.75-1.78|}{3} = \frac{|-0.26|+|-0.18|+|-0.03|}{3} \\ \\ = \frac{0.26+0.18+0.03}{3} = \frac{0.47}{3} \approx0.16$

Therefore, the mean absolute deviation </span><span>using exponential smoothing where α = 0.1 of October, November and December is given by: 0.157

</span><span><span>Part 5:

The mean absolute deviation of a forecast is given by the summation of the absolute values of the actual values minus the forecasted values all divided by the number of items.

Thus, given that the actual values of october, november and december are: $1.57, $1.62, $1.75

using </span><span>α = 0.3, we obtained that the forcasted values of october, november and december are: $1.76, $1.70, $1.68

Thus, the mean absolute deviation is given by:

$
 \frac{|1.57-1.76|+|1.62-1.70|+|1.75-1.68|}{3} = 
\frac{|-0.17|+|-0.08|+|-0.07|}{3} \\ \\ = \frac{0.17+0.08+0.07}{3} = 
\frac{0.32}{3} \approx0.107$

Therefore, the mean absolute deviation </span><span>using exponential smoothing where α = 0.3 of October, November and December is given by: 0.107

</span></span>
<span><span>Part 6:

The mean absolute deviation of a forecast is given by the summation of the absolute values of the actual values minus the forecasted values all divided by the number of items.

Thus, given that the actual values of october, november and december are: $1.57, $1.62, $1.75

using </span><span>α = 0.5, we obtained that the forcasted values of october, november and december are: $1.72, $1.65, $1.64

Thus, the mean absolute deviation is given by:

$
 \frac{|1.57-1.72|+|1.62-1.65|+|1.75-1.64|}{3} = 
\frac{|-0.15|+|-0.03|+|0.11|}{3} \\ \\ = \frac{0.15+0.03+0.11}{3} = 
\frac{29}{3} \approx0.097$

Therefore, the mean absolute deviation </span><span>using exponential smoothing where α = 0.5 of October, November and December is given by: 0.097</span></span>

5 0

2 years ago

5. Simplify the expressions below by combining like terms<br> a) 4 + x + 3x + 2x - 2

zalisa [80]

Answer:

6x + 2

Explanation:

4 + -2 = 2 and x + 3x + 2x = 6x

3 0

2 years ago

A square patio has an area of 200 square feet. How long is each side of the patio to the nearest tenth?

Darya [45]

If patio is square it means that area of if is $A=a^2$ where a is one side
Now we can substitute A=200 and solve eq
$a^2=200$
a= $\sqrt{200}$
a=14.1421
Now we are rounding to the nearest tenths
a=14.1 because next number is 4 and is less then 5.
The result is 14.1

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3 years ago

Luis earns a base salary of 150.00 every week with an additional 14% commission on everything he sells, luis sold 6050.00 worth

BlackZzzverrR [31]

I think It's $995 I HOPE THIS IS A CORRECT ANSWER

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2 years ago