How can I find values of $a,b,c$ when function is continuous?
$begingroup$
Given that the function $ fleft(xright) = left{
begin{array}{lr}
ax+2 & 0 le x le 1\
bx^2 +3ax +5 & 1<x le 2\
e^{2x} +b & 2 < x le 3\
(cx+3)^2 +2ax & 3 < x le 4\
end{array} right.\$
is continuous on the interval $[0,4]$. Find the values of $a,b $ and $ c$
I tried solving by equating right side limit of $1$ to the left side limit ($ax +2$ and $bx^2 + 3ax + 5$) but got stuck with equations.
limits continuity
edited Jan 13 at 15:23
KM101
5,9481524
asked Jan 13 at 15:14
user1444692user1444692
7416
$endgroup$
add a comment |
$begingroup$
Given that the function $ fleft(xright) = left{
begin{array}{lr}
ax+2 & 0 le x le 1\
bx^2 +3ax +5 & 1<x le 2\
e^{2x} +b & 2 < x le 3\
(cx+3)^2 +2ax & 3 < x le 4\
end{array} right.\$
is continuous on the interval $[0,4]$. Find the values of $a,b $ and $ c$
I tried solving by equating right side limit of $1$ to the left side limit ($ax +2$ and $bx^2 + 3ax + 5$) but got stuck with equations.
limits continuity
edited Jan 13 at 15:23
KM101
5,9481524
asked Jan 13 at 15:14
user1444692user1444692
7416
$endgroup$
add a comment |
$begingroup$
Given that the function $ fleft(xright) = left{
begin{array}{lr}
ax+2 & 0 le x le 1\
bx^2 +3ax +5 & 1<x le 2\
e^{2x} +b & 2 < x le 3\
(cx+3)^2 +2ax & 3 < x le 4\
end{array} right.\$
is continuous on the interval $[0,4]$. Find the values of $a,b $ and $ c$
I tried solving by equating right side limit of $1$ to the left side limit ($ax +2$ and $bx^2 + 3ax + 5$) but got stuck with equations.
limits continuity
edited Jan 13 at 15:23
KM101
5,9481524
asked Jan 13 at 15:14
user1444692user1444692
7416
$endgroup$
Given that the function $ fleft(xright) = left{
begin{array}{lr}
ax+2 & 0 le x le 1\
bx^2 +3ax +5 & 1<x le 2\
e^{2x} +b & 2 < x le 3\
(cx+3)^2 +2ax & 3 < x le 4\
end{array} right.\$
is continuous on the interval $[0,4]$. Find the values of $a,b $ and $ c$
I tried solving by equating right side limit of $1$ to the left side limit ($ax +2$ and $bx^2 + 3ax + 5$) but got stuck with equations.
limits continuity
limits continuity
edited Jan 13 at 15:23
KM101
5,9481524
asked Jan 13 at 15:14
user1444692user1444692
7416
edited Jan 13 at 15:23
KM101
5,9481524
asked Jan 13 at 15:14
user1444692user1444692
7416
edited Jan 13 at 15:23
KM101
5,9481524
edited Jan 13 at 15:23
KM101
5,9481524
edited Jan 13 at 15:23
KM101
5,9481524
5,9481524
asked Jan 13 at 15:14
user1444692user1444692
7416
asked Jan 13 at 15:14
user1444692user1444692
7416
asked Jan 13 at 15:14
user1444692user1444692
7416
7416
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Guide:
At $x=1$, you get $a(1)+2=b(1)^2+3a(1)+5$.
Similarly, you should be able to get equations at $x=2$ and $x=3$ respectively.
Now, solve the linear system of equations.
answered Jan 13 at 15:28
Siong Thye GohSiong Thye Goh
101k1466117
$endgroup$
$begingroup$
I tried but at x = 2, I got the same equation.
$endgroup$
– user1444692
Jan 13 at 15:29
$begingroup$
at $x=2$, we get $4b+6a+5=e^4+b$ which is very different from the first equation isn't it?
$endgroup$
– Siong Thye Goh
Jan 13 at 15:45
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Guide:
At $x=1$, you get $a(1)+2=b(1)^2+3a(1)+5$.
Similarly, you should be able to get equations at $x=2$ and $x=3$ respectively.
Now, solve the linear system of equations.
answered Jan 13 at 15:28
Siong Thye GohSiong Thye Goh
101k1466117
$endgroup$
$begingroup$
I tried but at x = 2, I got the same equation.
$endgroup$
– user1444692
Jan 13 at 15:29
$begingroup$
at $x=2$, we get $4b+6a+5=e^4+b$ which is very different from the first equation isn't it?
$endgroup$
– Siong Thye Goh
Jan 13 at 15:45
add a comment |
$begingroup$
Guide:
At $x=1$, you get $a(1)+2=b(1)^2+3a(1)+5$.
Similarly, you should be able to get equations at $x=2$ and $x=3$ respectively.
Now, solve the linear system of equations.
answered Jan 13 at 15:28
Siong Thye GohSiong Thye Goh
101k1466117
$endgroup$
$begingroup$
I tried but at x = 2, I got the same equation.
$endgroup$
– user1444692
Jan 13 at 15:29
$begingroup$
at $x=2$, we get $4b+6a+5=e^4+b$ which is very different from the first equation isn't it?
$endgroup$
– Siong Thye Goh
Jan 13 at 15:45
add a comment |
$begingroup$
Guide:
At $x=1$, you get $a(1)+2=b(1)^2+3a(1)+5$.
Similarly, you should be able to get equations at $x=2$ and $x=3$ respectively.
Now, solve the linear system of equations.
answered Jan 13 at 15:28
Siong Thye GohSiong Thye Goh
101k1466117
$endgroup$
Guide:
At $x=1$, you get $a(1)+2=b(1)^2+3a(1)+5$.
Similarly, you should be able to get equations at $x=2$ and $x=3$ respectively.
Now, solve the linear system of equations.
answered Jan 13 at 15:28
Siong Thye GohSiong Thye Goh
101k1466117
answered Jan 13 at 15:28
Siong Thye GohSiong Thye Goh
101k1466117
answered Jan 13 at 15:28
Siong Thye GohSiong Thye Goh
101k1466117
answered Jan 13 at 15:28
Siong Thye GohSiong Thye Goh
101k1466117
101k1466117
$begingroup$
I tried but at x = 2, I got the same equation.
$endgroup$
– user1444692
Jan 13 at 15:29
$begingroup$
at $x=2$, we get $4b+6a+5=e^4+b$ which is very different from the first equation isn't it?
$endgroup$
– Siong Thye Goh
Jan 13 at 15:45
add a comment |
$begingroup$
I tried but at x = 2, I got the same equation.
$endgroup$
– user1444692
Jan 13 at 15:29
$begingroup$
at $x=2$, we get $4b+6a+5=e^4+b$ which is very different from the first equation isn't it?
$endgroup$
– Siong Thye Goh
Jan 13 at 15:45
$begingroup$
I tried but at x = 2, I got the same equation.
$endgroup$
– user1444692
Jan 13 at 15:29
$begingroup$
I tried but at x = 2, I got the same equation.
$endgroup$
– user1444692
Jan 13 at 15:29
$begingroup$
at $x=2$, we get $4b+6a+5=e^4+b$ which is very different from the first equation isn't it?
$endgroup$
– Siong Thye Goh
Jan 13 at 15:45
$begingroup$
at $x=2$, we get $4b+6a+5=e^4+b$ which is very different from the first equation isn't it?
$endgroup$
– Siong Thye Goh
Jan 13 at 15:45
add a comment |
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