Mixing
let $f(x)=x^3-2x+4$ without finding them, explain why we know there must be values c1, c2, c3 $ldots$
$begingroup$
Let $f(x)=x^3-2x+4$ without finding them, explain why we know there must be values $c_1, c_2, c_3$ such that $f(c_1)=pi$, $f(c_2)=-sqrt 7$, $f(c_3)=1,000,000$ . I am not sure how to attack this question.
calculus
edited Feb 1 at 3:27
S.S.Danyal
5666
asked Feb 1 at 3:08
JamesJames
184
$endgroup$
add a comment |
$begingroup$
Let $f(x)=x^3-2x+4$ without finding them, explain why we know there must be values $c_1, c_2, c_3$ such that $f(c_1)=pi$, $f(c_2)=-sqrt 7$, $f(c_3)=1,000,000$ . I am not sure how to attack this question.
calculus
edited Feb 1 at 3:27
S.S.Danyal
5666
asked Feb 1 at 3:08
JamesJames
184
$endgroup$
add a comment |
$begingroup$
Let $f(x)=x^3-2x+4$ without finding them, explain why we know there must be values $c_1, c_2, c_3$ such that $f(c_1)=pi$, $f(c_2)=-sqrt 7$, $f(c_3)=1,000,000$ . I am not sure how to attack this question.
calculus
edited Feb 1 at 3:27
S.S.Danyal
5666
asked Feb 1 at 3:08
JamesJames
184
$endgroup$
Let $f(x)=x^3-2x+4$ without finding them, explain why we know there must be values $c_1, c_2, c_3$ such that $f(c_1)=pi$, $f(c_2)=-sqrt 7$, $f(c_3)=1,000,000$ . I am not sure how to attack this question.
calculus
calculus
edited Feb 1 at 3:27
S.S.Danyal
5666
asked Feb 1 at 3:08
JamesJames
184
edited Feb 1 at 3:27
S.S.Danyal
5666
asked Feb 1 at 3:08
JamesJames
184
edited Feb 1 at 3:27
S.S.Danyal
5666
edited Feb 1 at 3:27
S.S.Danyal
5666
edited Feb 1 at 3:27
S.S.Danyal
5666
5666
asked Feb 1 at 3:08
JamesJames
184
asked Feb 1 at 3:08
JamesJames
184
asked Feb 1 at 3:08
JamesJames
184
184
add a comment |
add a comment |
3 Answers
3
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oldest
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$begingroup$
Use Jimmy R's reasoning.
For example: Easy to check that $f(-1,000,000) < -2,000,000$ and $f(1,000,000) > 2,000,000$. Furthermore, $f$ is continuous. Thus as $1,000,000$ satisfies $f(-1,000,000) < -2,000,000 < 1,000,000 < 2,000,000 < f(1,000,000)$, the Intermediate Value Theorem says that there is an $x$ such that $f(x)$ is exactly $1,000,000$.
Finish the rest in the same fashion.
answered Feb 1 at 3:22
MikeMike
4,621512
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add a comment |
$begingroup$
Hint: Find the limits of $f(x)$ as $xtopminfty$. Then use that $f(x)$ is continuous, to infer that $f$ attains all the values between these limits.
answered Feb 1 at 3:12
Jimmy R.Jimmy R.
33.3k42257
$endgroup$
add a comment |
$begingroup$
Since $f$ is continuous, $f(mathbb R)$ is an intervall. We have $lim_{x to infty}f(x)=infty$ and $lim_{x to -infty}f(x)=-infty$, hence $f( mathbb R)= mathbb R.$
Can you take it from here ?
answered Feb 1 at 6:44
FredFred
48.5k11849
$endgroup$
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
$begingroup$
Use Jimmy R's reasoning.
For example: Easy to check that $f(-1,000,000) < -2,000,000$ and $f(1,000,000) > 2,000,000$. Furthermore, $f$ is continuous. Thus as $1,000,000$ satisfies $f(-1,000,000) < -2,000,000 < 1,000,000 < 2,000,000 < f(1,000,000)$, the Intermediate Value Theorem says that there is an $x$ such that $f(x)$ is exactly $1,000,000$.
Finish the rest in the same fashion.
answered Feb 1 at 3:22
MikeMike
4,621512
$endgroup$
add a comment |
$begingroup$
Use Jimmy R's reasoning.
For example: Easy to check that $f(-1,000,000) < -2,000,000$ and $f(1,000,000) > 2,000,000$. Furthermore, $f$ is continuous. Thus as $1,000,000$ satisfies $f(-1,000,000) < -2,000,000 < 1,000,000 < 2,000,000 < f(1,000,000)$, the Intermediate Value Theorem says that there is an $x$ such that $f(x)$ is exactly $1,000,000$.
Finish the rest in the same fashion.
answered Feb 1 at 3:22
MikeMike
4,621512
$endgroup$
add a comment |
$begingroup$
Use Jimmy R's reasoning.
For example: Easy to check that $f(-1,000,000) < -2,000,000$ and $f(1,000,000) > 2,000,000$. Furthermore, $f$ is continuous. Thus as $1,000,000$ satisfies $f(-1,000,000) < -2,000,000 < 1,000,000 < 2,000,000 < f(1,000,000)$, the Intermediate Value Theorem says that there is an $x$ such that $f(x)$ is exactly $1,000,000$.
Finish the rest in the same fashion.
answered Feb 1 at 3:22
MikeMike
4,621512
$endgroup$
Use Jimmy R's reasoning.
For example: Easy to check that $f(-1,000,000) < -2,000,000$ and $f(1,000,000) > 2,000,000$. Furthermore, $f$ is continuous. Thus as $1,000,000$ satisfies $f(-1,000,000) < -2,000,000 < 1,000,000 < 2,000,000 < f(1,000,000)$, the Intermediate Value Theorem says that there is an $x$ such that $f(x)$ is exactly $1,000,000$.
Finish the rest in the same fashion.
answered Feb 1 at 3:22
MikeMike
4,621512
answered Feb 1 at 3:22
MikeMike
4,621512
answered Feb 1 at 3:22
MikeMike
4,621512
answered Feb 1 at 3:22
MikeMike
4,621512
4,621512
add a comment |
add a comment |
$begingroup$
Hint: Find the limits of $f(x)$ as $xtopminfty$. Then use that $f(x)$ is continuous, to infer that $f$ attains all the values between these limits.
answered Feb 1 at 3:12
Jimmy R.Jimmy R.
33.3k42257
$endgroup$
add a comment |
$begingroup$
Hint: Find the limits of $f(x)$ as $xtopminfty$. Then use that $f(x)$ is continuous, to infer that $f$ attains all the values between these limits.
answered Feb 1 at 3:12
Jimmy R.Jimmy R.
33.3k42257
$endgroup$
add a comment |
$begingroup$
Hint: Find the limits of $f(x)$ as $xtopminfty$. Then use that $f(x)$ is continuous, to infer that $f$ attains all the values between these limits.
answered Feb 1 at 3:12
Jimmy R.Jimmy R.
33.3k42257
$endgroup$
Hint: Find the limits of $f(x)$ as $xtopminfty$. Then use that $f(x)$ is continuous, to infer that $f$ attains all the values between these limits.
answered Feb 1 at 3:12
Jimmy R.Jimmy R.
33.3k42257
answered Feb 1 at 3:12
Jimmy R.Jimmy R.
33.3k42257
answered Feb 1 at 3:12
Jimmy R.Jimmy R.
33.3k42257
answered Feb 1 at 3:12
Jimmy R.Jimmy R.
33.3k42257
33.3k42257
add a comment |
add a comment |
$begingroup$
Since $f$ is continuous, $f(mathbb R)$ is an intervall. We have $lim_{x to infty}f(x)=infty$ and $lim_{x to -infty}f(x)=-infty$, hence $f( mathbb R)= mathbb R.$
Can you take it from here ?
answered Feb 1 at 6:44
FredFred
48.5k11849
$endgroup$
add a comment |
$begingroup$
Since $f$ is continuous, $f(mathbb R)$ is an intervall. We have $lim_{x to infty}f(x)=infty$ and $lim_{x to -infty}f(x)=-infty$, hence $f( mathbb R)= mathbb R.$
Can you take it from here ?
answered Feb 1 at 6:44
FredFred
48.5k11849
$endgroup$
add a comment |
$begingroup$
Since $f$ is continuous, $f(mathbb R)$ is an intervall. We have $lim_{x to infty}f(x)=infty$ and $lim_{x to -infty}f(x)=-infty$, hence $f( mathbb R)= mathbb R.$
Can you take it from here ?
answered Feb 1 at 6:44
FredFred
48.5k11849
$endgroup$
Since $f$ is continuous, $f(mathbb R)$ is an intervall. We have $lim_{x to infty}f(x)=infty$ and $lim_{x to -infty}f(x)=-infty$, hence $f( mathbb R)= mathbb R.$
Can you take it from here ?
answered Feb 1 at 6:44
FredFred
48.5k11849
answered Feb 1 at 6:44
FredFred
48.5k11849
answered Feb 1 at 6:44
FredFred
48.5k11849
answered Feb 1 at 6:44
FredFred
48.5k11849
48.5k11849
add a comment |
add a comment |
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