Mixing
Finding a linear function from given functions
$begingroup$
The question is asking to find the linear function, $f(t) = vt + C$ for $f(t+2) = f(t) + 6$ and $f(1) = 10$
The answer is $3t + 7$, but I have no idea how the answer is produced.
functions polynomials
edited Jan 8 at 18:42
greedoid
40.7k1149100
asked Dec 27 '18 at 15:25
Evan KimEvan Kim
1068
$endgroup$
add a comment |
$begingroup$
The question is asking to find the linear function, $f(t) = vt + C$ for $f(t+2) = f(t) + 6$ and $f(1) = 10$
The answer is $3t + 7$, but I have no idea how the answer is produced.
functions polynomials
edited Jan 8 at 18:42
greedoid
40.7k1149100
asked Dec 27 '18 at 15:25
Evan KimEvan Kim
1068
$endgroup$
$begingroup$
Set up the two given conditions as equations to be solved for the coefficients $v$ and $C$.
$endgroup$
– zipirovich
Dec 27 '18 at 15:29
add a comment |
$begingroup$
The question is asking to find the linear function, $f(t) = vt + C$ for $f(t+2) = f(t) + 6$ and $f(1) = 10$
The answer is $3t + 7$, but I have no idea how the answer is produced.
functions polynomials
edited Jan 8 at 18:42
greedoid
40.7k1149100
asked Dec 27 '18 at 15:25
Evan KimEvan Kim
1068
$endgroup$
The question is asking to find the linear function, $f(t) = vt + C$ for $f(t+2) = f(t) + 6$ and $f(1) = 10$
The answer is $3t + 7$, but I have no idea how the answer is produced.
functions polynomials
functions polynomials
edited Jan 8 at 18:42
greedoid
40.7k1149100
asked Dec 27 '18 at 15:25
Evan KimEvan Kim
1068
edited Jan 8 at 18:42
greedoid
40.7k1149100
asked Dec 27 '18 at 15:25
Evan KimEvan Kim
1068
edited Jan 8 at 18:42
greedoid
40.7k1149100
edited Jan 8 at 18:42
greedoid
40.7k1149100
edited Jan 8 at 18:42
greedoid
40.7k1149100
40.7k1149100
asked Dec 27 '18 at 15:25
Evan KimEvan Kim
1068
asked Dec 27 '18 at 15:25
Evan KimEvan Kim
1068
asked Dec 27 '18 at 15:25
Evan KimEvan Kim
1068
1068
$begingroup$
Set up the two given conditions as equations to be solved for the coefficients $v$ and $C$.
$endgroup$
– zipirovich
Dec 27 '18 at 15:29
add a comment |
$begingroup$
Set up the two given conditions as equations to be solved for the coefficients $v$ and $C$.
$endgroup$
– zipirovich
Dec 27 '18 at 15:29
$begingroup$
Set up the two given conditions as equations to be solved for the coefficients $v$ and $C$.
$endgroup$
– zipirovich
Dec 27 '18 at 15:29
$begingroup$
Set up the two given conditions as equations to be solved for the coefficients $v$ and $C$.
$endgroup$
– zipirovich
Dec 27 '18 at 15:29
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Since $f(1)=10$ we have $f(3)= f(1)+6=16$. Thus you have to solve the system:
$$ v+c=10$$
$$ 3v+c = 16$$
answered Dec 27 '18 at 15:28
greedoidgreedoid
40.7k1149100
$endgroup$
add a comment |
$begingroup$
Hint:
$$f(t+2) = f(t)+6$$
$$f(1) = 10 = f(-1)+6 iff f(-1) = 4$$
You now have the two points $(1, 10)$ and $(-1, 4)$.
Small Addition: From here, you can either use the use a system of equations for the two points $(1, 10)$ and $(-1, 4)$:
$$begin{cases}
v+C = 10\
-v+C = 4
end{cases}$$
or you can refer to the definition of a linear equation $y = vt+C$, in which $v$ is the slope and $C$ is the $y$-intercept, or the $y$-coordinate when $x = 0$ (and it’s easy to find here because the point is the midpoint of the two points already found).
$$v = frac{Delta f(t)}{Delta t}$$
answered Dec 27 '18 at 15:29
KM101KM101
5,9481524
$endgroup$
add a comment |
$begingroup$
... and now the conceptual version. "$f(t+2) = f(t)+6$" means that the slope is $6$ (the increase in output) over a run of $2$ (the increase in input), so the slope is $frac{6}{2} = 3$. Then you have a point on the line, $(1,10)$ from "$f(1) = 10$", so $f(t) - 10 = 3(t-1)$, by point-slope, and we have $f(t) = 3t + 7$.
answered Dec 27 '18 at 22:23
Eric TowersEric Towers
32.5k22369
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Since $f(1)=10$ we have $f(3)= f(1)+6=16$. Thus you have to solve the system:
$$ v+c=10$$
$$ 3v+c = 16$$
answered Dec 27 '18 at 15:28
greedoidgreedoid
40.7k1149100
$endgroup$
add a comment |
$begingroup$
Since $f(1)=10$ we have $f(3)= f(1)+6=16$. Thus you have to solve the system:
$$ v+c=10$$
$$ 3v+c = 16$$
answered Dec 27 '18 at 15:28
greedoidgreedoid
40.7k1149100
$endgroup$
add a comment |
$begingroup$
Since $f(1)=10$ we have $f(3)= f(1)+6=16$. Thus you have to solve the system:
$$ v+c=10$$
$$ 3v+c = 16$$
answered Dec 27 '18 at 15:28
greedoidgreedoid
40.7k1149100
$endgroup$
Since $f(1)=10$ we have $f(3)= f(1)+6=16$. Thus you have to solve the system:
$$ v+c=10$$
$$ 3v+c = 16$$
answered Dec 27 '18 at 15:28
greedoidgreedoid
40.7k1149100
answered Dec 27 '18 at 15:28
greedoidgreedoid
40.7k1149100
answered Dec 27 '18 at 15:28
greedoidgreedoid
40.7k1149100
answered Dec 27 '18 at 15:28
greedoidgreedoid
40.7k1149100
40.7k1149100
add a comment |
add a comment |
$begingroup$
Hint:
$$f(t+2) = f(t)+6$$
$$f(1) = 10 = f(-1)+6 iff f(-1) = 4$$
You now have the two points $(1, 10)$ and $(-1, 4)$.
Small Addition: From here, you can either use the use a system of equations for the two points $(1, 10)$ and $(-1, 4)$:
$$begin{cases}
v+C = 10\
-v+C = 4
end{cases}$$
or you can refer to the definition of a linear equation $y = vt+C$, in which $v$ is the slope and $C$ is the $y$-intercept, or the $y$-coordinate when $x = 0$ (and it’s easy to find here because the point is the midpoint of the two points already found).
$$v = frac{Delta f(t)}{Delta t}$$
answered Dec 27 '18 at 15:29
KM101KM101
5,9481524
$endgroup$
add a comment |
$begingroup$
Hint:
$$f(t+2) = f(t)+6$$
$$f(1) = 10 = f(-1)+6 iff f(-1) = 4$$
You now have the two points $(1, 10)$ and $(-1, 4)$.
Small Addition: From here, you can either use the use a system of equations for the two points $(1, 10)$ and $(-1, 4)$:
$$begin{cases}
v+C = 10\
-v+C = 4
end{cases}$$
or you can refer to the definition of a linear equation $y = vt+C$, in which $v$ is the slope and $C$ is the $y$-intercept, or the $y$-coordinate when $x = 0$ (and it’s easy to find here because the point is the midpoint of the two points already found).
$$v = frac{Delta f(t)}{Delta t}$$
answered Dec 27 '18 at 15:29
KM101KM101
5,9481524
$endgroup$
add a comment |
$begingroup$
Hint:
$$f(t+2) = f(t)+6$$
$$f(1) = 10 = f(-1)+6 iff f(-1) = 4$$
You now have the two points $(1, 10)$ and $(-1, 4)$.
Small Addition: From here, you can either use the use a system of equations for the two points $(1, 10)$ and $(-1, 4)$:
$$begin{cases}
v+C = 10\
-v+C = 4
end{cases}$$
or you can refer to the definition of a linear equation $y = vt+C$, in which $v$ is the slope and $C$ is the $y$-intercept, or the $y$-coordinate when $x = 0$ (and it’s easy to find here because the point is the midpoint of the two points already found).
$$v = frac{Delta f(t)}{Delta t}$$
answered Dec 27 '18 at 15:29
KM101KM101
5,9481524
$endgroup$
Hint:
$$f(t+2) = f(t)+6$$
$$f(1) = 10 = f(-1)+6 iff f(-1) = 4$$
You now have the two points $(1, 10)$ and $(-1, 4)$.
Small Addition: From here, you can either use the use a system of equations for the two points $(1, 10)$ and $(-1, 4)$:
$$begin{cases}
v+C = 10\
-v+C = 4
end{cases}$$
or you can refer to the definition of a linear equation $y = vt+C$, in which $v$ is the slope and $C$ is the $y$-intercept, or the $y$-coordinate when $x = 0$ (and it’s easy to find here because the point is the midpoint of the two points already found).
$$v = frac{Delta f(t)}{Delta t}$$
answered Dec 27 '18 at 15:29
KM101KM101
5,9481524
edited Dec 27 '18 at 16:51
answered Dec 27 '18 at 15:29
KM101KM101
5,9481524
answered Dec 27 '18 at 15:29
KM101KM101
5,9481524
answered Dec 27 '18 at 15:29
KM101KM101
5,9481524
5,9481524
add a comment |
add a comment |
$begingroup$
... and now the conceptual version. "$f(t+2) = f(t)+6$" means that the slope is $6$ (the increase in output) over a run of $2$ (the increase in input), so the slope is $frac{6}{2} = 3$. Then you have a point on the line, $(1,10)$ from "$f(1) = 10$", so $f(t) - 10 = 3(t-1)$, by point-slope, and we have $f(t) = 3t + 7$.
answered Dec 27 '18 at 22:23
Eric TowersEric Towers
32.5k22369
$endgroup$
add a comment |
$begingroup$
... and now the conceptual version. "$f(t+2) = f(t)+6$" means that the slope is $6$ (the increase in output) over a run of $2$ (the increase in input), so the slope is $frac{6}{2} = 3$. Then you have a point on the line, $(1,10)$ from "$f(1) = 10$", so $f(t) - 10 = 3(t-1)$, by point-slope, and we have $f(t) = 3t + 7$.
answered Dec 27 '18 at 22:23
Eric TowersEric Towers
32.5k22369
$endgroup$
add a comment |
$begingroup$
... and now the conceptual version. "$f(t+2) = f(t)+6$" means that the slope is $6$ (the increase in output) over a run of $2$ (the increase in input), so the slope is $frac{6}{2} = 3$. Then you have a point on the line, $(1,10)$ from "$f(1) = 10$", so $f(t) - 10 = 3(t-1)$, by point-slope, and we have $f(t) = 3t + 7$.
answered Dec 27 '18 at 22:23
Eric TowersEric Towers
32.5k22369
$endgroup$
... and now the conceptual version. "$f(t+2) = f(t)+6$" means that the slope is $6$ (the increase in output) over a run of $2$ (the increase in input), so the slope is $frac{6}{2} = 3$. Then you have a point on the line, $(1,10)$ from "$f(1) = 10$", so $f(t) - 10 = 3(t-1)$, by point-slope, and we have $f(t) = 3t + 7$.
answered Dec 27 '18 at 22:23
Eric TowersEric Towers
32.5k22369
answered Dec 27 '18 at 22:23
Eric TowersEric Towers
32.5k22369
answered Dec 27 '18 at 22:23
Eric TowersEric Towers
32.5k22369
answered Dec 27 '18 at 22:23
Eric TowersEric Towers
32.5k22369
32.5k22369
add a comment |
add a comment |
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$begingroup$
Set up the two given conditions as equations to be solved for the coefficients $v$ and $C$.
$endgroup$
– zipirovich
Dec 27 '18 at 15:29