How do I use the Composite of Continuous Functions theorem to show that a function is continuous?
0
$$f(x)= sqrt frac{x}{x+1}$$
I know that I have to split it into two separate functions:
$$ g(x)= sqrt x$$
$$h(x)=frac{x}{x+1}$$
I'm just not sure what to do next, and I can't seem to find anything that explains it well enough (or simply enough, I guess). Am I just an idiot? Any help would really be appreciated!
calculus functions continuity function-and-relation-composition
asked Nov 28 '18 at 0:02
dumpster fire
1
add a comment |
0
$$f(x)= sqrt frac{x}{x+1}$$
I know that I have to split it into two separate functions:
$$ g(x)= sqrt x$$
$$h(x)=frac{x}{x+1}$$
I'm just not sure what to do next, and I can't seem to find anything that explains it well enough (or simply enough, I guess). Am I just an idiot? Any help would really be appreciated!
calculus functions continuity function-and-relation-composition
asked Nov 28 '18 at 0:02
dumpster fire
1
I would suggest you use the definition of continuity. In case you were just told "you can't pick up your pencil", try the definition that $lim_{x rightarrow x_0} f(x) = f(x_0)$.
– tekay-squared
Nov 28 '18 at 0:05
1
continuous on what domain?
– qbert
Nov 28 '18 at 0:06
$g$ and $h$ are continuous and $f(x) = g(h(x)$ (which you figured out). Then the theorem you quote says $f$ is continuous where it's defined- nothing more is required, unless you are supposed to find the domain of $f$ too. Can you figure out what values of $x$ will cause trouble?
– Ethan Bolker
Nov 28 '18 at 0:09
2
If you've already shown that $sqrt{x}$ is continuous then you're done the first step. Otherwise, prove it, and then also prove that $h$ is continuous. This can be done by recalling the theorem that says that the quotient of two continuous functions is continuous whenever the denominator is nonzero (notice that $g$ is just the quotient of $x$ and $x+1$, polynomials).
– AlkaKadri
Nov 28 '18 at 0:11
@EthanBolker Oh, that makes sense. Thank you!
– dumpster fire
Nov 28 '18 at 0:27
add a comment |
0
0
0
0
$$f(x)= sqrt frac{x}{x+1}$$
I know that I have to split it into two separate functions:
$$ g(x)= sqrt x$$
$$h(x)=frac{x}{x+1}$$
I'm just not sure what to do next, and I can't seem to find anything that explains it well enough (or simply enough, I guess). Am I just an idiot? Any help would really be appreciated!
calculus functions continuity function-and-relation-composition
asked Nov 28 '18 at 0:02
dumpster fire
1
$$f(x)= sqrt frac{x}{x+1}$$
I know that I have to split it into two separate functions:
$$ g(x)= sqrt x$$
$$h(x)=frac{x}{x+1}$$
I'm just not sure what to do next, and I can't seem to find anything that explains it well enough (or simply enough, I guess). Am I just an idiot? Any help would really be appreciated!
calculus functions continuity function-and-relation-composition
calculus functions continuity function-and-relation-composition
asked Nov 28 '18 at 0:02
dumpster fire
1
asked Nov 28 '18 at 0:02
dumpster fire
1
asked Nov 28 '18 at 0:02
dumpster fire
1
asked Nov 28 '18 at 0:02
dumpster fire
1
asked Nov 28 '18 at 0:02
dumpster fire
1
1
I would suggest you use the definition of continuity. In case you were just told "you can't pick up your pencil", try the definition that $lim_{x rightarrow x_0} f(x) = f(x_0)$.
– tekay-squared
Nov 28 '18 at 0:05
1
continuous on what domain?
– qbert
Nov 28 '18 at 0:06
$g$ and $h$ are continuous and $f(x) = g(h(x)$ (which you figured out). Then the theorem you quote says $f$ is continuous where it's defined- nothing more is required, unless you are supposed to find the domain of $f$ too. Can you figure out what values of $x$ will cause trouble?
– Ethan Bolker
Nov 28 '18 at 0:09
2
If you've already shown that $sqrt{x}$ is continuous then you're done the first step. Otherwise, prove it, and then also prove that $h$ is continuous. This can be done by recalling the theorem that says that the quotient of two continuous functions is continuous whenever the denominator is nonzero (notice that $g$ is just the quotient of $x$ and $x+1$, polynomials).
– AlkaKadri
Nov 28 '18 at 0:11
@EthanBolker Oh, that makes sense. Thank you!
– dumpster fire
Nov 28 '18 at 0:27
add a comment |
I would suggest you use the definition of continuity. In case you were just told "you can't pick up your pencil", try the definition that $lim_{x rightarrow x_0} f(x) = f(x_0)$.
– tekay-squared
Nov 28 '18 at 0:05
1
continuous on what domain?
– qbert
Nov 28 '18 at 0:06
$g$ and $h$ are continuous and $f(x) = g(h(x)$ (which you figured out). Then the theorem you quote says $f$ is continuous where it's defined- nothing more is required, unless you are supposed to find the domain of $f$ too. Can you figure out what values of $x$ will cause trouble?
– Ethan Bolker
Nov 28 '18 at 0:09
2
If you've already shown that $sqrt{x}$ is continuous then you're done the first step. Otherwise, prove it, and then also prove that $h$ is continuous. This can be done by recalling the theorem that says that the quotient of two continuous functions is continuous whenever the denominator is nonzero (notice that $g$ is just the quotient of $x$ and $x+1$, polynomials).
– AlkaKadri
Nov 28 '18 at 0:11
@EthanBolker Oh, that makes sense. Thank you!
– dumpster fire
Nov 28 '18 at 0:27
I would suggest you use the definition of continuity. In case you were just told "you can't pick up your pencil", try the definition that $lim_{x rightarrow x_0} f(x) = f(x_0)$.
– tekay-squared
Nov 28 '18 at 0:05
I would suggest you use the definition of continuity. In case you were just told "you can't pick up your pencil", try the definition that $lim_{x rightarrow x_0} f(x) = f(x_0)$.
– tekay-squared
Nov 28 '18 at 0:05
1
1
continuous on what domain?
– qbert
Nov 28 '18 at 0:06
continuous on what domain?
– qbert
Nov 28 '18 at 0:06
$g$ and $h$ are continuous and $f(x) = g(h(x)$ (which you figured out). Then the theorem you quote says $f$ is continuous where it's defined- nothing more is required, unless you are supposed to find the domain of $f$ too. Can you figure out what values of $x$ will cause trouble?
– Ethan Bolker
Nov 28 '18 at 0:09
$g$ and $h$ are continuous and $f(x) = g(h(x)$ (which you figured out). Then the theorem you quote says $f$ is continuous where it's defined- nothing more is required, unless you are supposed to find the domain of $f$ too. Can you figure out what values of $x$ will cause trouble?
– Ethan Bolker
Nov 28 '18 at 0:09
2
2
If you've already shown that $sqrt{x}$ is continuous then you're done the first step. Otherwise, prove it, and then also prove that $h$ is continuous. This can be done by recalling the theorem that says that the quotient of two continuous functions is continuous whenever the denominator is nonzero (notice that $g$ is just the quotient of $x$ and $x+1$, polynomials).
– AlkaKadri
Nov 28 '18 at 0:11
If you've already shown that $sqrt{x}$ is continuous then you're done the first step. Otherwise, prove it, and then also prove that $h$ is continuous. This can be done by recalling the theorem that says that the quotient of two continuous functions is continuous whenever the denominator is nonzero (notice that $g$ is just the quotient of $x$ and $x+1$, polynomials).
– AlkaKadri
Nov 28 '18 at 0:11
@EthanBolker Oh, that makes sense. Thank you!
– dumpster fire
Nov 28 '18 at 0:27
@EthanBolker Oh, that makes sense. Thank you!
– dumpster fire
Nov 28 '18 at 0:27
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I would suggest you use the definition of continuity. In case you were just told "you can't pick up your pencil", try the definition that $lim_{x rightarrow x_0} f(x) = f(x_0)$.
– tekay-squared
Nov 28 '18 at 0:05
1
continuous on what domain?
– qbert
Nov 28 '18 at 0:06
$g$ and $h$ are continuous and $f(x) = g(h(x)$ (which you figured out). Then the theorem you quote says $f$ is continuous where it's defined- nothing more is required, unless you are supposed to find the domain of $f$ too. Can you figure out what values of $x$ will cause trouble?
– Ethan Bolker
Nov 28 '18 at 0:09
2
If you've already shown that $sqrt{x}$ is continuous then you're done the first step. Otherwise, prove it, and then also prove that $h$ is continuous. This can be done by recalling the theorem that says that the quotient of two continuous functions is continuous whenever the denominator is nonzero (notice that $g$ is just the quotient of $x$ and $x+1$, polynomials).
– AlkaKadri
Nov 28 '18 at 0:11
@EthanBolker Oh, that makes sense. Thank you!
– dumpster fire
Nov 28 '18 at 0:27