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f: [0,1]→R be a function differentiable in (0,1) such that f(0) = 1 and f′(x)≥3 for all x∈(0,1). [closed]
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Let
$f: [0,1]→mathbb R$ be a function differentiable in $(0,1)$ such that $f(0) = 1$ and $f′(x)≥3$ for all $x∈(0,1)$. What is the smallest possible value that such a function can take at $x=1/2$
calculus
edited Aug 1 at 0:35
user529760
509216
asked Jul 31 at 23:58
boxerbiggles
92
closed as off-topic by amWhy, Rory Daulton, Arnaud Mortier, Leucippus, Xander Henderson Aug 1 at 2:21
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Rory Daulton, Arnaud Mortier, Leucippus, Xander Henderson
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up vote
-3
down vote
favorite
Let
$f: [0,1]→mathbb R$ be a function differentiable in $(0,1)$ such that $f(0) = 1$ and $f′(x)≥3$ for all $x∈(0,1)$. What is the smallest possible value that such a function can take at $x=1/2$
calculus
edited Aug 1 at 0:35
user529760
509216
asked Jul 31 at 23:58
boxerbiggles
92
closed as off-topic by amWhy, Rory Daulton, Arnaud Mortier, Leucippus, Xander Henderson Aug 1 at 2:21
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Rory Daulton, Arnaud Mortier, Leucippus, Xander Henderson
You are more likely to get a good answer to your question if you follow a few guidelines. In particular, what have you tried so far, and just where are you stuck? This is not a homework-answering site: many of us want to see that you have put significant work into the problem.
– Rory Daulton
Aug 1 at 0:08
Assume that $f(1/2)<3/2+1$, then $fracf(1/2)-f(0)1/2-0=fracf(1/2)-11/2<3$. But by the Mean Value Theorem there must be a point $cin(0,1/2)$ such that $f'(c)=fracf(1/2)-f(0)1/2-0<3$. That contradicts the assumption that $f'(x)geq3$ for all $xin(0,1)$. Therefore, $f(1/2)geq 3/2+1$. This value can be attained for the function $f(x)=3x+1$.
– JessicaMcRae
Aug 1 at 0:29
Since you have $f'(x)≥3$. Integrate both sides , you will get $f(x)≥3x+c$ , then plug in $x=0$ to get the value of $c$ which will be $1$. Hence $f(x)≥3x+1$. Plug in $x=1/2$ and evaluate the minimum possible value of $f(1/2)$ which will be $5/2$.
– Alphanerd
Aug 1 at 5:37
add a comment |Â
up vote
-3
down vote
favorite
up vote
-3
down vote
favorite
Let
$f: [0,1]→mathbb R$ be a function differentiable in $(0,1)$ such that $f(0) = 1$ and $f′(x)≥3$ for all $x∈(0,1)$. What is the smallest possible value that such a function can take at $x=1/2$
calculus
edited Aug 1 at 0:35
user529760
509216
asked Jul 31 at 23:58
boxerbiggles
92
Let
$f: [0,1]→mathbb R$ be a function differentiable in $(0,1)$ such that $f(0) = 1$ and $f′(x)≥3$ for all $x∈(0,1)$. What is the smallest possible value that such a function can take at $x=1/2$
calculus
edited Aug 1 at 0:35
user529760
509216
asked Jul 31 at 23:58
boxerbiggles
92
edited Aug 1 at 0:35
user529760
509216
edited Aug 1 at 0:35
user529760
509216
edited Aug 1 at 0:35
user529760
509216
509216
asked Jul 31 at 23:58
boxerbiggles
92
asked Jul 31 at 23:58
boxerbiggles
92
asked Jul 31 at 23:58
boxerbiggles
92
92
closed as off-topic by amWhy, Rory Daulton, Arnaud Mortier, Leucippus, Xander Henderson Aug 1 at 2:21
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Rory Daulton, Arnaud Mortier, Leucippus, Xander Henderson
closed as off-topic by amWhy, Rory Daulton, Arnaud Mortier, Leucippus, Xander Henderson Aug 1 at 2:21
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Rory Daulton, Arnaud Mortier, Leucippus, Xander Henderson
You are more likely to get a good answer to your question if you follow a few guidelines. In particular, what have you tried so far, and just where are you stuck? This is not a homework-answering site: many of us want to see that you have put significant work into the problem.
– Rory Daulton
Aug 1 at 0:08
Assume that $f(1/2)<3/2+1$, then $fracf(1/2)-f(0)1/2-0=fracf(1/2)-11/2<3$. But by the Mean Value Theorem there must be a point $cin(0,1/2)$ such that $f'(c)=fracf(1/2)-f(0)1/2-0<3$. That contradicts the assumption that $f'(x)geq3$ for all $xin(0,1)$. Therefore, $f(1/2)geq 3/2+1$. This value can be attained for the function $f(x)=3x+1$.
– JessicaMcRae
Aug 1 at 0:29
Since you have $f'(x)≥3$. Integrate both sides , you will get $f(x)≥3x+c$ , then plug in $x=0$ to get the value of $c$ which will be $1$. Hence $f(x)≥3x+1$. Plug in $x=1/2$ and evaluate the minimum possible value of $f(1/2)$ which will be $5/2$.
– Alphanerd
Aug 1 at 5:37
add a comment |Â
You are more likely to get a good answer to your question if you follow a few guidelines. In particular, what have you tried so far, and just where are you stuck? This is not a homework-answering site: many of us want to see that you have put significant work into the problem.
– Rory Daulton
Aug 1 at 0:08
Assume that $f(1/2)<3/2+1$, then $fracf(1/2)-f(0)1/2-0=fracf(1/2)-11/2<3$. But by the Mean Value Theorem there must be a point $cin(0,1/2)$ such that $f'(c)=fracf(1/2)-f(0)1/2-0<3$. That contradicts the assumption that $f'(x)geq3$ for all $xin(0,1)$. Therefore, $f(1/2)geq 3/2+1$. This value can be attained for the function $f(x)=3x+1$.
– JessicaMcRae
Aug 1 at 0:29
Since you have $f'(x)≥3$. Integrate both sides , you will get $f(x)≥3x+c$ , then plug in $x=0$ to get the value of $c$ which will be $1$. Hence $f(x)≥3x+1$. Plug in $x=1/2$ and evaluate the minimum possible value of $f(1/2)$ which will be $5/2$.
– Alphanerd
Aug 1 at 5:37
You are more likely to get a good answer to your question if you follow a few guidelines. In particular, what have you tried so far, and just where are you stuck? This is not a homework-answering site: many of us want to see that you have put significant work into the problem.
– Rory Daulton
Aug 1 at 0:08
You are more likely to get a good answer to your question if you follow a few guidelines. In particular, what have you tried so far, and just where are you stuck? This is not a homework-answering site: many of us want to see that you have put significant work into the problem.
– Rory Daulton
Aug 1 at 0:08
Assume that $f(1/2)<3/2+1$, then $fracf(1/2)-f(0)1/2-0=fracf(1/2)-11/2<3$. But by the Mean Value Theorem there must be a point $cin(0,1/2)$ such that $f'(c)=fracf(1/2)-f(0)1/2-0<3$. That contradicts the assumption that $f'(x)geq3$ for all $xin(0,1)$. Therefore, $f(1/2)geq 3/2+1$. This value can be attained for the function $f(x)=3x+1$.
– JessicaMcRae
Aug 1 at 0:29
Assume that $f(1/2)<3/2+1$, then $fracf(1/2)-f(0)1/2-0=fracf(1/2)-11/2<3$. But by the Mean Value Theorem there must be a point $cin(0,1/2)$ such that $f'(c)=fracf(1/2)-f(0)1/2-0<3$. That contradicts the assumption that $f'(x)geq3$ for all $xin(0,1)$. Therefore, $f(1/2)geq 3/2+1$. This value can be attained for the function $f(x)=3x+1$.
– JessicaMcRae
Aug 1 at 0:29
Since you have $f'(x)≥3$. Integrate both sides , you will get $f(x)≥3x+c$ , then plug in $x=0$ to get the value of $c$ which will be $1$. Hence $f(x)≥3x+1$. Plug in $x=1/2$ and evaluate the minimum possible value of $f(1/2)$ which will be $5/2$.
– Alphanerd
Aug 1 at 5:37
Since you have $f'(x)≥3$. Integrate both sides , you will get $f(x)≥3x+c$ , then plug in $x=0$ to get the value of $c$ which will be $1$. Hence $f(x)≥3x+1$. Plug in $x=1/2$ and evaluate the minimum possible value of $f(1/2)$ which will be $5/2$.
– Alphanerd
Aug 1 at 5:37
add a comment |Â
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You are more likely to get a good answer to your question if you follow a few guidelines. In particular, what have you tried so far, and just where are you stuck? This is not a homework-answering site: many of us want to see that you have put significant work into the problem.
– Rory Daulton
Aug 1 at 0:08
Assume that $f(1/2)<3/2+1$, then $fracf(1/2)-f(0)1/2-0=fracf(1/2)-11/2<3$. But by the Mean Value Theorem there must be a point $cin(0,1/2)$ such that $f'(c)=fracf(1/2)-f(0)1/2-0<3$. That contradicts the assumption that $f'(x)geq3$ for all $xin(0,1)$. Therefore, $f(1/2)geq 3/2+1$. This value can be attained for the function $f(x)=3x+1$.
– JessicaMcRae
Aug 1 at 0:29
Since you have $f'(x)≥3$. Integrate both sides , you will get $f(x)≥3x+c$ , then plug in $x=0$ to get the value of $c$ which will be $1$. Hence $f(x)≥3x+1$. Plug in $x=1/2$ and evaluate the minimum possible value of $f(1/2)$ which will be $5/2$.
– Alphanerd
Aug 1 at 5:37