Mixing
Need some help with continuity for the following piece-wise functions
Let
$$f(x)=begin{cases}x^2−7 &xleq c\10x−32 &x>c end{cases}$$
If $f(x)$ is continuous everywhere, then what is $c$ equal to?
Also, where is $$f(x)=begin{cases}x+2 &x<0\e^x &0leq xleq 1\ 2−x &x>1end{cases}$$
continuous?
calculus continuity piecewise-continuity
edited Nov 21 '18 at 6:20
Yadati Kiran
1,693619
asked Nov 21 '18 at 0:31
sgill0
12
add a comment |
Let
$$f(x)=begin{cases}x^2−7 &xleq c\10x−32 &x>c end{cases}$$
If $f(x)$ is continuous everywhere, then what is $c$ equal to?
Also, where is $$f(x)=begin{cases}x+2 &x<0\e^x &0leq xleq 1\ 2−x &x>1end{cases}$$
continuous?
calculus continuity piecewise-continuity
edited Nov 21 '18 at 6:20
Yadati Kiran
1,693619
asked Nov 21 '18 at 0:31
sgill0
12
What have you tried? Are you stuck somewhere? We are suppose to help, not solve it for you
– Andrei
Nov 21 '18 at 0:34
I have no idea where to begin with the first question. For the second function, I have tried to graph it and it seems as though the function is continuous from (-infinity,infinity)
– sgill0
Nov 21 '18 at 0:38
add a comment |
Let
$$f(x)=begin{cases}x^2−7 &xleq c\10x−32 &x>c end{cases}$$
If $f(x)$ is continuous everywhere, then what is $c$ equal to?
Also, where is $$f(x)=begin{cases}x+2 &x<0\e^x &0leq xleq 1\ 2−x &x>1end{cases}$$
continuous?
calculus continuity piecewise-continuity
edited Nov 21 '18 at 6:20
Yadati Kiran
1,693619
asked Nov 21 '18 at 0:31
sgill0
12
Let
$$f(x)=begin{cases}x^2−7 &xleq c\10x−32 &x>c end{cases}$$
If $f(x)$ is continuous everywhere, then what is $c$ equal to?
Also, where is $$f(x)=begin{cases}x+2 &x<0\e^x &0leq xleq 1\ 2−x &x>1end{cases}$$
continuous?
calculus continuity piecewise-continuity
calculus continuity piecewise-continuity
edited Nov 21 '18 at 6:20
Yadati Kiran
1,693619
asked Nov 21 '18 at 0:31
sgill0
12
edited Nov 21 '18 at 6:20
Yadati Kiran
1,693619
asked Nov 21 '18 at 0:31
sgill0
12
edited Nov 21 '18 at 6:20
Yadati Kiran
1,693619
edited Nov 21 '18 at 6:20
Yadati Kiran
1,693619
edited Nov 21 '18 at 6:20
Yadati Kiran
1,693619
1,693619
asked Nov 21 '18 at 0:31
sgill0
12
asked Nov 21 '18 at 0:31
sgill0
12
asked Nov 21 '18 at 0:31
sgill0
12
12
What have you tried? Are you stuck somewhere? We are suppose to help, not solve it for you
– Andrei
Nov 21 '18 at 0:34
I have no idea where to begin with the first question. For the second function, I have tried to graph it and it seems as though the function is continuous from (-infinity,infinity)
– sgill0
Nov 21 '18 at 0:38
add a comment |
What have you tried? Are you stuck somewhere? We are suppose to help, not solve it for you
– Andrei
Nov 21 '18 at 0:34
I have no idea where to begin with the first question. For the second function, I have tried to graph it and it seems as though the function is continuous from (-infinity,infinity)
– sgill0
Nov 21 '18 at 0:38
What have you tried? Are you stuck somewhere? We are suppose to help, not solve it for you
– Andrei
Nov 21 '18 at 0:34
What have you tried? Are you stuck somewhere? We are suppose to help, not solve it for you
– Andrei
Nov 21 '18 at 0:34
I have no idea where to begin with the first question. For the second function, I have tried to graph it and it seems as though the function is continuous from (-infinity,infinity)
– sgill0
Nov 21 '18 at 0:38
I have no idea where to begin with the first question. For the second function, I have tried to graph it and it seems as though the function is continuous from (-infinity,infinity)
– sgill0
Nov 21 '18 at 0:38
add a comment |
3 Answers
3
active
oldest
votes
Hint:
- Use the following for both parts, if a function $f$ is continous at $c$, then we have
$$lim_{x to c^-}f(x) = f(c) = lim_{x to c^+}f(x)$$
- Exponential and polynomials are continuous.
answered Nov 21 '18 at 0:50
Siong Thye Goh
99.5k1465117
add a comment |
Since $f$ is continuous everywhere, it must be continuous particularly at $c$ as well. So $displaystyle lim_{x to c^-}f(x)=lim_{x to c^+}implies c^2-7=10c-32implies (c-5)^2=0$ or $c=5$. For the second part, check for continuity at the intersection points $Big($i.e. $displaystyle lim_{x to c^-}f(x)=lim_{x to c^+}$$Big)$ of the intervals since polynomial and exponential functions are continuous. So at $x=0$ we have $0+2neq e^0=1$ and at $x=1$ we have $e^1neq2-1$. Hence the interval of convergence of $f$ is $(-infty,0)cup(0,1)cup(0,infty)$
answered Nov 21 '18 at 0:55
Yadati Kiran
1,693619
add a comment |
For the first question it's clear that each of those pieces of that function are continuous on their entire respective domains. Therefore, you just need to make sure it's continuous at the point where you switch between the two, at $x=c$. For a function to be continuous at $x=c$ you need $$lim_{x to c}f(x)=f(c).$$ Since $f(x)$ is different from the left and right of $x=c$, you need to show
$$lim_{x to c^{-}} f(x) = f(c)$$
$$lim_{x to c^{+}} f(x) = f(c).$$
The left sided limit would give you
$$lim_{x to c^{-}} (x^2-7) = c^2-7$$
Which would actually be true for any $c$ and is not very useful. However, the right sided limit would give you
$$lim_{x to c^{+}} (10x-32) = c^2-7.$$
Solving this equation for $c$ would tell you what value(s) for $c$ that would make the function continuous everywhere.
Here's a link to a site with a bit more detailed answer to a similar problem. It goes a bit more into why this works and why you want to solve it this way:
https://jakesmathlessons.com/limits/solution-find-the-values-of-a-and-b-that-make-f-continuous-everywhere/
The second question will be similar. All three of those pieces of that function are continuous on their own. As a result, you can say that $f(x)$ is continuous for all $x<0$, $0<x<1$, and $x>1$. You just need to show whether it is also continuous at $x=0$ and $x=1$. You will want to do this just like how I made sure the previous problem is continuous at $x=c$. The only difference is that you will get equations that don't have variables. If the equations are true then it's continuous at that point, if it's false then the function isn't continuous at that point.
For example, to show continuity at $x=0$, you need to show
$$lim_{x to 0^-}f(x)=f(0)$$
and
$$lim_{x to 0^+}f(x)=f(0)$$
The right sided limit is not useful, but the left sided limit gives:
$$lim_{x to 0^-}(x+2)=e^0$$
$$2=1$$
Since we know this is not true, $f(x)$ must not be continuous at $x=0$. Now do the same thing with $x=1$ to see if it's continuous there or not.
answered Nov 22 '18 at 0:28
Jake O
514
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
Hint:
- Use the following for both parts, if a function $f$ is continous at $c$, then we have
$$lim_{x to c^-}f(x) = f(c) = lim_{x to c^+}f(x)$$
- Exponential and polynomials are continuous.
answered Nov 21 '18 at 0:50
Siong Thye Goh
99.5k1465117
add a comment |
Hint:
- Use the following for both parts, if a function $f$ is continous at $c$, then we have
$$lim_{x to c^-}f(x) = f(c) = lim_{x to c^+}f(x)$$
- Exponential and polynomials are continuous.
answered Nov 21 '18 at 0:50
Siong Thye Goh
99.5k1465117
add a comment |
Hint:
- Use the following for both parts, if a function $f$ is continous at $c$, then we have
$$lim_{x to c^-}f(x) = f(c) = lim_{x to c^+}f(x)$$
- Exponential and polynomials are continuous.
answered Nov 21 '18 at 0:50
Siong Thye Goh
99.5k1465117
Hint:
- Use the following for both parts, if a function $f$ is continous at $c$, then we have
$$lim_{x to c^-}f(x) = f(c) = lim_{x to c^+}f(x)$$
- Exponential and polynomials are continuous.
answered Nov 21 '18 at 0:50
Siong Thye Goh
99.5k1465117
answered Nov 21 '18 at 0:50
Siong Thye Goh
99.5k1465117
answered Nov 21 '18 at 0:50
Siong Thye Goh
99.5k1465117
answered Nov 21 '18 at 0:50
Siong Thye Goh
99.5k1465117
99.5k1465117
add a comment |
add a comment |
Since $f$ is continuous everywhere, it must be continuous particularly at $c$ as well. So $displaystyle lim_{x to c^-}f(x)=lim_{x to c^+}implies c^2-7=10c-32implies (c-5)^2=0$ or $c=5$. For the second part, check for continuity at the intersection points $Big($i.e. $displaystyle lim_{x to c^-}f(x)=lim_{x to c^+}$$Big)$ of the intervals since polynomial and exponential functions are continuous. So at $x=0$ we have $0+2neq e^0=1$ and at $x=1$ we have $e^1neq2-1$. Hence the interval of convergence of $f$ is $(-infty,0)cup(0,1)cup(0,infty)$
answered Nov 21 '18 at 0:55
Yadati Kiran
1,693619
add a comment |
Since $f$ is continuous everywhere, it must be continuous particularly at $c$ as well. So $displaystyle lim_{x to c^-}f(x)=lim_{x to c^+}implies c^2-7=10c-32implies (c-5)^2=0$ or $c=5$. For the second part, check for continuity at the intersection points $Big($i.e. $displaystyle lim_{x to c^-}f(x)=lim_{x to c^+}$$Big)$ of the intervals since polynomial and exponential functions are continuous. So at $x=0$ we have $0+2neq e^0=1$ and at $x=1$ we have $e^1neq2-1$. Hence the interval of convergence of $f$ is $(-infty,0)cup(0,1)cup(0,infty)$
answered Nov 21 '18 at 0:55
Yadati Kiran
1,693619
add a comment |
Since $f$ is continuous everywhere, it must be continuous particularly at $c$ as well. So $displaystyle lim_{x to c^-}f(x)=lim_{x to c^+}implies c^2-7=10c-32implies (c-5)^2=0$ or $c=5$. For the second part, check for continuity at the intersection points $Big($i.e. $displaystyle lim_{x to c^-}f(x)=lim_{x to c^+}$$Big)$ of the intervals since polynomial and exponential functions are continuous. So at $x=0$ we have $0+2neq e^0=1$ and at $x=1$ we have $e^1neq2-1$. Hence the interval of convergence of $f$ is $(-infty,0)cup(0,1)cup(0,infty)$
answered Nov 21 '18 at 0:55
Yadati Kiran
1,693619
Since $f$ is continuous everywhere, it must be continuous particularly at $c$ as well. So $displaystyle lim_{x to c^-}f(x)=lim_{x to c^+}implies c^2-7=10c-32implies (c-5)^2=0$ or $c=5$. For the second part, check for continuity at the intersection points $Big($i.e. $displaystyle lim_{x to c^-}f(x)=lim_{x to c^+}$$Big)$ of the intervals since polynomial and exponential functions are continuous. So at $x=0$ we have $0+2neq e^0=1$ and at $x=1$ we have $e^1neq2-1$. Hence the interval of convergence of $f$ is $(-infty,0)cup(0,1)cup(0,infty)$
answered Nov 21 '18 at 0:55
Yadati Kiran
1,693619
answered Nov 21 '18 at 0:55
Yadati Kiran
1,693619
answered Nov 21 '18 at 0:55
Yadati Kiran
1,693619
answered Nov 21 '18 at 0:55
Yadati Kiran
1,693619
1,693619
add a comment |
add a comment |
For the first question it's clear that each of those pieces of that function are continuous on their entire respective domains. Therefore, you just need to make sure it's continuous at the point where you switch between the two, at $x=c$. For a function to be continuous at $x=c$ you need $$lim_{x to c}f(x)=f(c).$$ Since $f(x)$ is different from the left and right of $x=c$, you need to show
$$lim_{x to c^{-}} f(x) = f(c)$$
$$lim_{x to c^{+}} f(x) = f(c).$$
The left sided limit would give you
$$lim_{x to c^{-}} (x^2-7) = c^2-7$$
Which would actually be true for any $c$ and is not very useful. However, the right sided limit would give you
$$lim_{x to c^{+}} (10x-32) = c^2-7.$$
Solving this equation for $c$ would tell you what value(s) for $c$ that would make the function continuous everywhere.
Here's a link to a site with a bit more detailed answer to a similar problem. It goes a bit more into why this works and why you want to solve it this way:
https://jakesmathlessons.com/limits/solution-find-the-values-of-a-and-b-that-make-f-continuous-everywhere/
The second question will be similar. All three of those pieces of that function are continuous on their own. As a result, you can say that $f(x)$ is continuous for all $x<0$, $0<x<1$, and $x>1$. You just need to show whether it is also continuous at $x=0$ and $x=1$. You will want to do this just like how I made sure the previous problem is continuous at $x=c$. The only difference is that you will get equations that don't have variables. If the equations are true then it's continuous at that point, if it's false then the function isn't continuous at that point.
For example, to show continuity at $x=0$, you need to show
$$lim_{x to 0^-}f(x)=f(0)$$
and
$$lim_{x to 0^+}f(x)=f(0)$$
The right sided limit is not useful, but the left sided limit gives:
$$lim_{x to 0^-}(x+2)=e^0$$
$$2=1$$
Since we know this is not true, $f(x)$ must not be continuous at $x=0$. Now do the same thing with $x=1$ to see if it's continuous there or not.
answered Nov 22 '18 at 0:28
Jake O
514
add a comment |
For the first question it's clear that each of those pieces of that function are continuous on their entire respective domains. Therefore, you just need to make sure it's continuous at the point where you switch between the two, at $x=c$. For a function to be continuous at $x=c$ you need $$lim_{x to c}f(x)=f(c).$$ Since $f(x)$ is different from the left and right of $x=c$, you need to show
$$lim_{x to c^{-}} f(x) = f(c)$$
$$lim_{x to c^{+}} f(x) = f(c).$$
The left sided limit would give you
$$lim_{x to c^{-}} (x^2-7) = c^2-7$$
Which would actually be true for any $c$ and is not very useful. However, the right sided limit would give you
$$lim_{x to c^{+}} (10x-32) = c^2-7.$$
Solving this equation for $c$ would tell you what value(s) for $c$ that would make the function continuous everywhere.
Here's a link to a site with a bit more detailed answer to a similar problem. It goes a bit more into why this works and why you want to solve it this way:
https://jakesmathlessons.com/limits/solution-find-the-values-of-a-and-b-that-make-f-continuous-everywhere/
The second question will be similar. All three of those pieces of that function are continuous on their own. As a result, you can say that $f(x)$ is continuous for all $x<0$, $0<x<1$, and $x>1$. You just need to show whether it is also continuous at $x=0$ and $x=1$. You will want to do this just like how I made sure the previous problem is continuous at $x=c$. The only difference is that you will get equations that don't have variables. If the equations are true then it's continuous at that point, if it's false then the function isn't continuous at that point.
For example, to show continuity at $x=0$, you need to show
$$lim_{x to 0^-}f(x)=f(0)$$
and
$$lim_{x to 0^+}f(x)=f(0)$$
The right sided limit is not useful, but the left sided limit gives:
$$lim_{x to 0^-}(x+2)=e^0$$
$$2=1$$
Since we know this is not true, $f(x)$ must not be continuous at $x=0$. Now do the same thing with $x=1$ to see if it's continuous there or not.
answered Nov 22 '18 at 0:28
Jake O
514
add a comment |
For the first question it's clear that each of those pieces of that function are continuous on their entire respective domains. Therefore, you just need to make sure it's continuous at the point where you switch between the two, at $x=c$. For a function to be continuous at $x=c$ you need $$lim_{x to c}f(x)=f(c).$$ Since $f(x)$ is different from the left and right of $x=c$, you need to show
$$lim_{x to c^{-}} f(x) = f(c)$$
$$lim_{x to c^{+}} f(x) = f(c).$$
The left sided limit would give you
$$lim_{x to c^{-}} (x^2-7) = c^2-7$$
Which would actually be true for any $c$ and is not very useful. However, the right sided limit would give you
$$lim_{x to c^{+}} (10x-32) = c^2-7.$$
Solving this equation for $c$ would tell you what value(s) for $c$ that would make the function continuous everywhere.
Here's a link to a site with a bit more detailed answer to a similar problem. It goes a bit more into why this works and why you want to solve it this way:
https://jakesmathlessons.com/limits/solution-find-the-values-of-a-and-b-that-make-f-continuous-everywhere/
The second question will be similar. All three of those pieces of that function are continuous on their own. As a result, you can say that $f(x)$ is continuous for all $x<0$, $0<x<1$, and $x>1$. You just need to show whether it is also continuous at $x=0$ and $x=1$. You will want to do this just like how I made sure the previous problem is continuous at $x=c$. The only difference is that you will get equations that don't have variables. If the equations are true then it's continuous at that point, if it's false then the function isn't continuous at that point.
For example, to show continuity at $x=0$, you need to show
$$lim_{x to 0^-}f(x)=f(0)$$
and
$$lim_{x to 0^+}f(x)=f(0)$$
The right sided limit is not useful, but the left sided limit gives:
$$lim_{x to 0^-}(x+2)=e^0$$
$$2=1$$
Since we know this is not true, $f(x)$ must not be continuous at $x=0$. Now do the same thing with $x=1$ to see if it's continuous there or not.
answered Nov 22 '18 at 0:28
Jake O
514
For the first question it's clear that each of those pieces of that function are continuous on their entire respective domains. Therefore, you just need to make sure it's continuous at the point where you switch between the two, at $x=c$. For a function to be continuous at $x=c$ you need $$lim_{x to c}f(x)=f(c).$$ Since $f(x)$ is different from the left and right of $x=c$, you need to show
$$lim_{x to c^{-}} f(x) = f(c)$$
$$lim_{x to c^{+}} f(x) = f(c).$$
The left sided limit would give you
$$lim_{x to c^{-}} (x^2-7) = c^2-7$$
Which would actually be true for any $c$ and is not very useful. However, the right sided limit would give you
$$lim_{x to c^{+}} (10x-32) = c^2-7.$$
Solving this equation for $c$ would tell you what value(s) for $c$ that would make the function continuous everywhere.
Here's a link to a site with a bit more detailed answer to a similar problem. It goes a bit more into why this works and why you want to solve it this way:
https://jakesmathlessons.com/limits/solution-find-the-values-of-a-and-b-that-make-f-continuous-everywhere/
The second question will be similar. All three of those pieces of that function are continuous on their own. As a result, you can say that $f(x)$ is continuous for all $x<0$, $0<x<1$, and $x>1$. You just need to show whether it is also continuous at $x=0$ and $x=1$. You will want to do this just like how I made sure the previous problem is continuous at $x=c$. The only difference is that you will get equations that don't have variables. If the equations are true then it's continuous at that point, if it's false then the function isn't continuous at that point.
For example, to show continuity at $x=0$, you need to show
$$lim_{x to 0^-}f(x)=f(0)$$
and
$$lim_{x to 0^+}f(x)=f(0)$$
The right sided limit is not useful, but the left sided limit gives:
$$lim_{x to 0^-}(x+2)=e^0$$
$$2=1$$
Since we know this is not true, $f(x)$ must not be continuous at $x=0$. Now do the same thing with $x=1$ to see if it's continuous there or not.
answered Nov 22 '18 at 0:28
Jake O
514
edited Nov 22 '18 at 0:39
answered Nov 22 '18 at 0:28
Jake O
514
answered Nov 22 '18 at 0:28
Jake O
514
answered Nov 22 '18 at 0:28
Jake O
514
514
add a comment |
add a comment |
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What have you tried? Are you stuck somewhere? We are suppose to help, not solve it for you
– Andrei
Nov 21 '18 at 0:34
I have no idea where to begin with the first question. For the second function, I have tried to graph it and it seems as though the function is continuous from (-infinity,infinity)
– sgill0
Nov 21 '18 at 0:38