Find values of $a$ if $f$ is a one-one function
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If $$f(x) = x^3 + (a+2)x^2 + 5ax + 5$$ is a one-one function then find the set of values of $a$.
I know that I need to find values of $a$ for which function is strictly monotonic increasing or strictly monotonic decreasing and we check monotonicity of function by its derivative $f'(x)$.
here $$f'(x) = 3x^2 + 2x(a+2) + 5a$$
so, what condition should I impose on $f'(x)$ to make the given funtion one one or strictly monotonic?
calculus
$endgroup$
add a comment |
$begingroup$
If $$f(x) = x^3 + (a+2)x^2 + 5ax + 5$$ is a one-one function then find the set of values of $a$.
I know that I need to find values of $a$ for which function is strictly monotonic increasing or strictly monotonic decreasing and we check monotonicity of function by its derivative $f'(x)$.
here $$f'(x) = 3x^2 + 2x(a+2) + 5a$$
so, what condition should I impose on $f'(x)$ to make the given funtion one one or strictly monotonic?
calculus
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3
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You need only pick $a $ so that $f'(x)$ does not have two zeros. Use its discriminant.
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– coffeemath
Dec 3 '18 at 5:04
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Note: It's OK if $f'$ has only one zero. Then $f$ will still be strictly increasing or strictly decreaing, even though it is flat at a single point. [By strictly increasing is meant if $a<b$ then $f(a)<f(b).$ Similar for strictly decreasing.]
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– coffeemath
Dec 3 '18 at 5:53
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@coffeemath this is what i thought. Now, f is parabola, then it can be one one only if x takes positive real values. So, is it okay to take domain (0,infinity)?
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– Mathsaddict
Dec 3 '18 at 6:07
1
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No, look at formula for $f$ -- it's cubic so $f$ will not be a parabola. Its derivative will be. And if that derivative doesn't cross the axis twice, then the original $f$ will be 1-1 for all values of $x$ [i.e.domain all reals]
$endgroup$
– coffeemath
Dec 3 '18 at 6:20
add a comment |
$begingroup$
If $$f(x) = x^3 + (a+2)x^2 + 5ax + 5$$ is a one-one function then find the set of values of $a$.
I know that I need to find values of $a$ for which function is strictly monotonic increasing or strictly monotonic decreasing and we check monotonicity of function by its derivative $f'(x)$.
here $$f'(x) = 3x^2 + 2x(a+2) + 5a$$
so, what condition should I impose on $f'(x)$ to make the given funtion one one or strictly monotonic?
calculus
$endgroup$
If $$f(x) = x^3 + (a+2)x^2 + 5ax + 5$$ is a one-one function then find the set of values of $a$.
I know that I need to find values of $a$ for which function is strictly monotonic increasing or strictly monotonic decreasing and we check monotonicity of function by its derivative $f'(x)$.
here $$f'(x) = 3x^2 + 2x(a+2) + 5a$$
so, what condition should I impose on $f'(x)$ to make the given funtion one one or strictly monotonic?
calculus
calculus
edited Dec 3 '18 at 5:05
Tianlalu
3,08621038
3,08621038
asked Dec 3 '18 at 4:56
MathsaddictMathsaddict
3049
3049
3
$begingroup$
You need only pick $a $ so that $f'(x)$ does not have two zeros. Use its discriminant.
$endgroup$
– coffeemath
Dec 3 '18 at 5:04
$begingroup$
Note: It's OK if $f'$ has only one zero. Then $f$ will still be strictly increasing or strictly decreaing, even though it is flat at a single point. [By strictly increasing is meant if $a<b$ then $f(a)<f(b).$ Similar for strictly decreasing.]
$endgroup$
– coffeemath
Dec 3 '18 at 5:53
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@coffeemath this is what i thought. Now, f is parabola, then it can be one one only if x takes positive real values. So, is it okay to take domain (0,infinity)?
$endgroup$
– Mathsaddict
Dec 3 '18 at 6:07
1
$begingroup$
No, look at formula for $f$ -- it's cubic so $f$ will not be a parabola. Its derivative will be. And if that derivative doesn't cross the axis twice, then the original $f$ will be 1-1 for all values of $x$ [i.e.domain all reals]
$endgroup$
– coffeemath
Dec 3 '18 at 6:20
add a comment |
3
$begingroup$
You need only pick $a $ so that $f'(x)$ does not have two zeros. Use its discriminant.
$endgroup$
– coffeemath
Dec 3 '18 at 5:04
$begingroup$
Note: It's OK if $f'$ has only one zero. Then $f$ will still be strictly increasing or strictly decreaing, even though it is flat at a single point. [By strictly increasing is meant if $a<b$ then $f(a)<f(b).$ Similar for strictly decreasing.]
$endgroup$
– coffeemath
Dec 3 '18 at 5:53
$begingroup$
@coffeemath this is what i thought. Now, f is parabola, then it can be one one only if x takes positive real values. So, is it okay to take domain (0,infinity)?
$endgroup$
– Mathsaddict
Dec 3 '18 at 6:07
1
$begingroup$
No, look at formula for $f$ -- it's cubic so $f$ will not be a parabola. Its derivative will be. And if that derivative doesn't cross the axis twice, then the original $f$ will be 1-1 for all values of $x$ [i.e.domain all reals]
$endgroup$
– coffeemath
Dec 3 '18 at 6:20
3
3
$begingroup$
You need only pick $a $ so that $f'(x)$ does not have two zeros. Use its discriminant.
$endgroup$
– coffeemath
Dec 3 '18 at 5:04
$begingroup$
You need only pick $a $ so that $f'(x)$ does not have two zeros. Use its discriminant.
$endgroup$
– coffeemath
Dec 3 '18 at 5:04
$begingroup$
Note: It's OK if $f'$ has only one zero. Then $f$ will still be strictly increasing or strictly decreaing, even though it is flat at a single point. [By strictly increasing is meant if $a<b$ then $f(a)<f(b).$ Similar for strictly decreasing.]
$endgroup$
– coffeemath
Dec 3 '18 at 5:53
$begingroup$
Note: It's OK if $f'$ has only one zero. Then $f$ will still be strictly increasing or strictly decreaing, even though it is flat at a single point. [By strictly increasing is meant if $a<b$ then $f(a)<f(b).$ Similar for strictly decreasing.]
$endgroup$
– coffeemath
Dec 3 '18 at 5:53
$begingroup$
@coffeemath this is what i thought. Now, f is parabola, then it can be one one only if x takes positive real values. So, is it okay to take domain (0,infinity)?
$endgroup$
– Mathsaddict
Dec 3 '18 at 6:07
$begingroup$
@coffeemath this is what i thought. Now, f is parabola, then it can be one one only if x takes positive real values. So, is it okay to take domain (0,infinity)?
$endgroup$
– Mathsaddict
Dec 3 '18 at 6:07
1
1
$begingroup$
No, look at formula for $f$ -- it's cubic so $f$ will not be a parabola. Its derivative will be. And if that derivative doesn't cross the axis twice, then the original $f$ will be 1-1 for all values of $x$ [i.e.domain all reals]
$endgroup$
– coffeemath
Dec 3 '18 at 6:20
$begingroup$
No, look at formula for $f$ -- it's cubic so $f$ will not be a parabola. Its derivative will be. And if that derivative doesn't cross the axis twice, then the original $f$ will be 1-1 for all values of $x$ [i.e.domain all reals]
$endgroup$
– coffeemath
Dec 3 '18 at 6:20
add a comment |
4 Answers
4
active
oldest
votes
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HINT. You know that $f'(x)$ is a parabola. To make $f(x)$ always increasing, you want $f'(x)>0$; that is, you want the parabola to always be above the $x$-axis. It touches the $x$-axis when it only has one root. What formula can you use to see what makes this only have one root in terms of $a$? Would this then give you a condition on $a$ to force the parabola to be always above the $x$-axis?
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Thanks for this simple explanation. Since, f is a parabola, so is there also any condition on x? I think domain of f should be positive real numbers, since f is one one.
$endgroup$
– Mathsaddict
Dec 3 '18 at 6:00
$begingroup$
@Mathsaddict there's no restriction on f taking any real number. The way I interpreted f being "one-to-one" is f is a bijection from R to itself.
$endgroup$
– Vim
Dec 3 '18 at 6:15
add a comment |
$begingroup$
A differentiable function is strictly monotonic when its derivative is not $0$ on any segment, meaning if $f'(x)=0$ then there is a small $z>0$ with the property that for any $0<y<z :f'(x+y)neq 0,f'(x-y)neq 0$, and if $forall x:f'(x)geq 0$ or $forall x:f'(x)leq 0$.
Try and use that. You are on the right track.
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add a comment |
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Consider $y=ax^2+bx+c,$ where $a>0.$ Now $$y=a(x^2+frac{b}{a}x+frac{c}{a})=abig((x+frac{b}{2a})^2+frac{c}{a}-frac{b^2}{4a^2}big)=abig((x+frac{b}{2a})^2+frac{4ac-b^2}{4a^2}big).$$ So if you have $4ac-b^2geq0$ then $y$ will be non-negative for all $x.$
Try to connect these ideas to your problem.
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if 4ac - b^2 = 0, then y will be zero at -b/2a. There should be no problem if f' has only one root. Why do i need 4ac - b^2 to be greater then 0? I think it can also be 0 at one point.
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– Mathsaddict
Dec 3 '18 at 5:58
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@Mathsaddict I agree with your comment here. [see my second comment on your question]t
$endgroup$
– coffeemath
Dec 3 '18 at 6:03
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@coffeemath I agree with yall. Edited.
$endgroup$
– Janitha357
Dec 3 '18 at 15:48
add a comment |
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Note that $lim_limits{xto -infty} f(x)=-infty$ and $lim_limits{xto +infty} f(x)=+infty$. That is why you need to make sure the function $f(x)$ is strictly increasing $left(x_1<x_2 Rightarrow f(x_1)<f(x_2)right)$. It implies:
$$f'(x)ge 0 Rightarrow 3x^2 + 2x(a+2) + 5age 0 Rightarrow Dle 0 Rightarrow (a+2)^2-15ale0 Rightarrow \
frac12 left(11 - sqrt{105}right)le alefrac12 left(11 + sqrt{105}right).$$
See WA animation for $f'(x)$.
Also note that $f'(x)=0$ when $a=frac12 left(11 - sqrt{105}right) left(text{or} a=frac12 left(11 + sqrt{105}right) right)$ and $x = frac{sqrt{35/3}}{2} - frac52 left(text{or} x = -frac52 - frac{sqrt{35/3}}{2}right)$, which implies it is an inflection point (which is still one-to-one).
See WA animation for $f(x)$.
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1
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Why D cannot be equal to 0?
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– Mathsaddict
Dec 3 '18 at 6:50
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Thank you for asking critical question. Please, see the updated answer.
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– farruhota
Dec 3 '18 at 15:55
add a comment |
Your Answer
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
HINT. You know that $f'(x)$ is a parabola. To make $f(x)$ always increasing, you want $f'(x)>0$; that is, you want the parabola to always be above the $x$-axis. It touches the $x$-axis when it only has one root. What formula can you use to see what makes this only have one root in terms of $a$? Would this then give you a condition on $a$ to force the parabola to be always above the $x$-axis?
$endgroup$
$begingroup$
Thanks for this simple explanation. Since, f is a parabola, so is there also any condition on x? I think domain of f should be positive real numbers, since f is one one.
$endgroup$
– Mathsaddict
Dec 3 '18 at 6:00
$begingroup$
@Mathsaddict there's no restriction on f taking any real number. The way I interpreted f being "one-to-one" is f is a bijection from R to itself.
$endgroup$
– Vim
Dec 3 '18 at 6:15
add a comment |
$begingroup$
HINT. You know that $f'(x)$ is a parabola. To make $f(x)$ always increasing, you want $f'(x)>0$; that is, you want the parabola to always be above the $x$-axis. It touches the $x$-axis when it only has one root. What formula can you use to see what makes this only have one root in terms of $a$? Would this then give you a condition on $a$ to force the parabola to be always above the $x$-axis?
$endgroup$
$begingroup$
Thanks for this simple explanation. Since, f is a parabola, so is there also any condition on x? I think domain of f should be positive real numbers, since f is one one.
$endgroup$
– Mathsaddict
Dec 3 '18 at 6:00
$begingroup$
@Mathsaddict there's no restriction on f taking any real number. The way I interpreted f being "one-to-one" is f is a bijection from R to itself.
$endgroup$
– Vim
Dec 3 '18 at 6:15
add a comment |
$begingroup$
HINT. You know that $f'(x)$ is a parabola. To make $f(x)$ always increasing, you want $f'(x)>0$; that is, you want the parabola to always be above the $x$-axis. It touches the $x$-axis when it only has one root. What formula can you use to see what makes this only have one root in terms of $a$? Would this then give you a condition on $a$ to force the parabola to be always above the $x$-axis?
$endgroup$
HINT. You know that $f'(x)$ is a parabola. To make $f(x)$ always increasing, you want $f'(x)>0$; that is, you want the parabola to always be above the $x$-axis. It touches the $x$-axis when it only has one root. What formula can you use to see what makes this only have one root in terms of $a$? Would this then give you a condition on $a$ to force the parabola to be always above the $x$-axis?
answered Dec 3 '18 at 5:05
mathematics2x2lifemathematics2x2life
8,06121738
8,06121738
$begingroup$
Thanks for this simple explanation. Since, f is a parabola, so is there also any condition on x? I think domain of f should be positive real numbers, since f is one one.
$endgroup$
– Mathsaddict
Dec 3 '18 at 6:00
$begingroup$
@Mathsaddict there's no restriction on f taking any real number. The way I interpreted f being "one-to-one" is f is a bijection from R to itself.
$endgroup$
– Vim
Dec 3 '18 at 6:15
add a comment |
$begingroup$
Thanks for this simple explanation. Since, f is a parabola, so is there also any condition on x? I think domain of f should be positive real numbers, since f is one one.
$endgroup$
– Mathsaddict
Dec 3 '18 at 6:00
$begingroup$
@Mathsaddict there's no restriction on f taking any real number. The way I interpreted f being "one-to-one" is f is a bijection from R to itself.
$endgroup$
– Vim
Dec 3 '18 at 6:15
$begingroup$
Thanks for this simple explanation. Since, f is a parabola, so is there also any condition on x? I think domain of f should be positive real numbers, since f is one one.
$endgroup$
– Mathsaddict
Dec 3 '18 at 6:00
$begingroup$
Thanks for this simple explanation. Since, f is a parabola, so is there also any condition on x? I think domain of f should be positive real numbers, since f is one one.
$endgroup$
– Mathsaddict
Dec 3 '18 at 6:00
$begingroup$
@Mathsaddict there's no restriction on f taking any real number. The way I interpreted f being "one-to-one" is f is a bijection from R to itself.
$endgroup$
– Vim
Dec 3 '18 at 6:15
$begingroup$
@Mathsaddict there's no restriction on f taking any real number. The way I interpreted f being "one-to-one" is f is a bijection from R to itself.
$endgroup$
– Vim
Dec 3 '18 at 6:15
add a comment |
$begingroup$
A differentiable function is strictly monotonic when its derivative is not $0$ on any segment, meaning if $f'(x)=0$ then there is a small $z>0$ with the property that for any $0<y<z :f'(x+y)neq 0,f'(x-y)neq 0$, and if $forall x:f'(x)geq 0$ or $forall x:f'(x)leq 0$.
Try and use that. You are on the right track.
$endgroup$
add a comment |
$begingroup$
A differentiable function is strictly monotonic when its derivative is not $0$ on any segment, meaning if $f'(x)=0$ then there is a small $z>0$ with the property that for any $0<y<z :f'(x+y)neq 0,f'(x-y)neq 0$, and if $forall x:f'(x)geq 0$ or $forall x:f'(x)leq 0$.
Try and use that. You are on the right track.
$endgroup$
add a comment |
$begingroup$
A differentiable function is strictly monotonic when its derivative is not $0$ on any segment, meaning if $f'(x)=0$ then there is a small $z>0$ with the property that for any $0<y<z :f'(x+y)neq 0,f'(x-y)neq 0$, and if $forall x:f'(x)geq 0$ or $forall x:f'(x)leq 0$.
Try and use that. You are on the right track.
$endgroup$
A differentiable function is strictly monotonic when its derivative is not $0$ on any segment, meaning if $f'(x)=0$ then there is a small $z>0$ with the property that for any $0<y<z :f'(x+y)neq 0,f'(x-y)neq 0$, and if $forall x:f'(x)geq 0$ or $forall x:f'(x)leq 0$.
Try and use that. You are on the right track.
answered Dec 3 '18 at 5:05
NL1992NL1992
7311
7311
add a comment |
add a comment |
$begingroup$
Consider $y=ax^2+bx+c,$ where $a>0.$ Now $$y=a(x^2+frac{b}{a}x+frac{c}{a})=abig((x+frac{b}{2a})^2+frac{c}{a}-frac{b^2}{4a^2}big)=abig((x+frac{b}{2a})^2+frac{4ac-b^2}{4a^2}big).$$ So if you have $4ac-b^2geq0$ then $y$ will be non-negative for all $x.$
Try to connect these ideas to your problem.
$endgroup$
$begingroup$
if 4ac - b^2 = 0, then y will be zero at -b/2a. There should be no problem if f' has only one root. Why do i need 4ac - b^2 to be greater then 0? I think it can also be 0 at one point.
$endgroup$
– Mathsaddict
Dec 3 '18 at 5:58
$begingroup$
@Mathsaddict I agree with your comment here. [see my second comment on your question]t
$endgroup$
– coffeemath
Dec 3 '18 at 6:03
$begingroup$
@coffeemath I agree with yall. Edited.
$endgroup$
– Janitha357
Dec 3 '18 at 15:48
add a comment |
$begingroup$
Consider $y=ax^2+bx+c,$ where $a>0.$ Now $$y=a(x^2+frac{b}{a}x+frac{c}{a})=abig((x+frac{b}{2a})^2+frac{c}{a}-frac{b^2}{4a^2}big)=abig((x+frac{b}{2a})^2+frac{4ac-b^2}{4a^2}big).$$ So if you have $4ac-b^2geq0$ then $y$ will be non-negative for all $x.$
Try to connect these ideas to your problem.
$endgroup$
$begingroup$
if 4ac - b^2 = 0, then y will be zero at -b/2a. There should be no problem if f' has only one root. Why do i need 4ac - b^2 to be greater then 0? I think it can also be 0 at one point.
$endgroup$
– Mathsaddict
Dec 3 '18 at 5:58
$begingroup$
@Mathsaddict I agree with your comment here. [see my second comment on your question]t
$endgroup$
– coffeemath
Dec 3 '18 at 6:03
$begingroup$
@coffeemath I agree with yall. Edited.
$endgroup$
– Janitha357
Dec 3 '18 at 15:48
add a comment |
$begingroup$
Consider $y=ax^2+bx+c,$ where $a>0.$ Now $$y=a(x^2+frac{b}{a}x+frac{c}{a})=abig((x+frac{b}{2a})^2+frac{c}{a}-frac{b^2}{4a^2}big)=abig((x+frac{b}{2a})^2+frac{4ac-b^2}{4a^2}big).$$ So if you have $4ac-b^2geq0$ then $y$ will be non-negative for all $x.$
Try to connect these ideas to your problem.
$endgroup$
Consider $y=ax^2+bx+c,$ where $a>0.$ Now $$y=a(x^2+frac{b}{a}x+frac{c}{a})=abig((x+frac{b}{2a})^2+frac{c}{a}-frac{b^2}{4a^2}big)=abig((x+frac{b}{2a})^2+frac{4ac-b^2}{4a^2}big).$$ So if you have $4ac-b^2geq0$ then $y$ will be non-negative for all $x.$
Try to connect these ideas to your problem.
edited Dec 3 '18 at 15:48
answered Dec 3 '18 at 5:23
Janitha357Janitha357
1,756516
1,756516
$begingroup$
if 4ac - b^2 = 0, then y will be zero at -b/2a. There should be no problem if f' has only one root. Why do i need 4ac - b^2 to be greater then 0? I think it can also be 0 at one point.
$endgroup$
– Mathsaddict
Dec 3 '18 at 5:58
$begingroup$
@Mathsaddict I agree with your comment here. [see my second comment on your question]t
$endgroup$
– coffeemath
Dec 3 '18 at 6:03
$begingroup$
@coffeemath I agree with yall. Edited.
$endgroup$
– Janitha357
Dec 3 '18 at 15:48
add a comment |
$begingroup$
if 4ac - b^2 = 0, then y will be zero at -b/2a. There should be no problem if f' has only one root. Why do i need 4ac - b^2 to be greater then 0? I think it can also be 0 at one point.
$endgroup$
– Mathsaddict
Dec 3 '18 at 5:58
$begingroup$
@Mathsaddict I agree with your comment here. [see my second comment on your question]t
$endgroup$
– coffeemath
Dec 3 '18 at 6:03
$begingroup$
@coffeemath I agree with yall. Edited.
$endgroup$
– Janitha357
Dec 3 '18 at 15:48
$begingroup$
if 4ac - b^2 = 0, then y will be zero at -b/2a. There should be no problem if f' has only one root. Why do i need 4ac - b^2 to be greater then 0? I think it can also be 0 at one point.
$endgroup$
– Mathsaddict
Dec 3 '18 at 5:58
$begingroup$
if 4ac - b^2 = 0, then y will be zero at -b/2a. There should be no problem if f' has only one root. Why do i need 4ac - b^2 to be greater then 0? I think it can also be 0 at one point.
$endgroup$
– Mathsaddict
Dec 3 '18 at 5:58
$begingroup$
@Mathsaddict I agree with your comment here. [see my second comment on your question]t
$endgroup$
– coffeemath
Dec 3 '18 at 6:03
$begingroup$
@Mathsaddict I agree with your comment here. [see my second comment on your question]t
$endgroup$
– coffeemath
Dec 3 '18 at 6:03
$begingroup$
@coffeemath I agree with yall. Edited.
$endgroup$
– Janitha357
Dec 3 '18 at 15:48
$begingroup$
@coffeemath I agree with yall. Edited.
$endgroup$
– Janitha357
Dec 3 '18 at 15:48
add a comment |
$begingroup$
Note that $lim_limits{xto -infty} f(x)=-infty$ and $lim_limits{xto +infty} f(x)=+infty$. That is why you need to make sure the function $f(x)$ is strictly increasing $left(x_1<x_2 Rightarrow f(x_1)<f(x_2)right)$. It implies:
$$f'(x)ge 0 Rightarrow 3x^2 + 2x(a+2) + 5age 0 Rightarrow Dle 0 Rightarrow (a+2)^2-15ale0 Rightarrow \
frac12 left(11 - sqrt{105}right)le alefrac12 left(11 + sqrt{105}right).$$
See WA animation for $f'(x)$.
Also note that $f'(x)=0$ when $a=frac12 left(11 - sqrt{105}right) left(text{or} a=frac12 left(11 + sqrt{105}right) right)$ and $x = frac{sqrt{35/3}}{2} - frac52 left(text{or} x = -frac52 - frac{sqrt{35/3}}{2}right)$, which implies it is an inflection point (which is still one-to-one).
See WA animation for $f(x)$.
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1
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Why D cannot be equal to 0?
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– Mathsaddict
Dec 3 '18 at 6:50
$begingroup$
Thank you for asking critical question. Please, see the updated answer.
$endgroup$
– farruhota
Dec 3 '18 at 15:55
add a comment |
$begingroup$
Note that $lim_limits{xto -infty} f(x)=-infty$ and $lim_limits{xto +infty} f(x)=+infty$. That is why you need to make sure the function $f(x)$ is strictly increasing $left(x_1<x_2 Rightarrow f(x_1)<f(x_2)right)$. It implies:
$$f'(x)ge 0 Rightarrow 3x^2 + 2x(a+2) + 5age 0 Rightarrow Dle 0 Rightarrow (a+2)^2-15ale0 Rightarrow \
frac12 left(11 - sqrt{105}right)le alefrac12 left(11 + sqrt{105}right).$$
See WA animation for $f'(x)$.
Also note that $f'(x)=0$ when $a=frac12 left(11 - sqrt{105}right) left(text{or} a=frac12 left(11 + sqrt{105}right) right)$ and $x = frac{sqrt{35/3}}{2} - frac52 left(text{or} x = -frac52 - frac{sqrt{35/3}}{2}right)$, which implies it is an inflection point (which is still one-to-one).
See WA animation for $f(x)$.
$endgroup$
1
$begingroup$
Why D cannot be equal to 0?
$endgroup$
– Mathsaddict
Dec 3 '18 at 6:50
$begingroup$
Thank you for asking critical question. Please, see the updated answer.
$endgroup$
– farruhota
Dec 3 '18 at 15:55
add a comment |
$begingroup$
Note that $lim_limits{xto -infty} f(x)=-infty$ and $lim_limits{xto +infty} f(x)=+infty$. That is why you need to make sure the function $f(x)$ is strictly increasing $left(x_1<x_2 Rightarrow f(x_1)<f(x_2)right)$. It implies:
$$f'(x)ge 0 Rightarrow 3x^2 + 2x(a+2) + 5age 0 Rightarrow Dle 0 Rightarrow (a+2)^2-15ale0 Rightarrow \
frac12 left(11 - sqrt{105}right)le alefrac12 left(11 + sqrt{105}right).$$
See WA animation for $f'(x)$.
Also note that $f'(x)=0$ when $a=frac12 left(11 - sqrt{105}right) left(text{or} a=frac12 left(11 + sqrt{105}right) right)$ and $x = frac{sqrt{35/3}}{2} - frac52 left(text{or} x = -frac52 - frac{sqrt{35/3}}{2}right)$, which implies it is an inflection point (which is still one-to-one).
See WA animation for $f(x)$.
$endgroup$
Note that $lim_limits{xto -infty} f(x)=-infty$ and $lim_limits{xto +infty} f(x)=+infty$. That is why you need to make sure the function $f(x)$ is strictly increasing $left(x_1<x_2 Rightarrow f(x_1)<f(x_2)right)$. It implies:
$$f'(x)ge 0 Rightarrow 3x^2 + 2x(a+2) + 5age 0 Rightarrow Dle 0 Rightarrow (a+2)^2-15ale0 Rightarrow \
frac12 left(11 - sqrt{105}right)le alefrac12 left(11 + sqrt{105}right).$$
See WA animation for $f'(x)$.
Also note that $f'(x)=0$ when $a=frac12 left(11 - sqrt{105}right) left(text{or} a=frac12 left(11 + sqrt{105}right) right)$ and $x = frac{sqrt{35/3}}{2} - frac52 left(text{or} x = -frac52 - frac{sqrt{35/3}}{2}right)$, which implies it is an inflection point (which is still one-to-one).
See WA animation for $f(x)$.
edited Dec 3 '18 at 15:54
answered Dec 3 '18 at 6:33
farruhotafarruhota
19.8k2738
19.8k2738
1
$begingroup$
Why D cannot be equal to 0?
$endgroup$
– Mathsaddict
Dec 3 '18 at 6:50
$begingroup$
Thank you for asking critical question. Please, see the updated answer.
$endgroup$
– farruhota
Dec 3 '18 at 15:55
add a comment |
1
$begingroup$
Why D cannot be equal to 0?
$endgroup$
– Mathsaddict
Dec 3 '18 at 6:50
$begingroup$
Thank you for asking critical question. Please, see the updated answer.
$endgroup$
– farruhota
Dec 3 '18 at 15:55
1
1
$begingroup$
Why D cannot be equal to 0?
$endgroup$
– Mathsaddict
Dec 3 '18 at 6:50
$begingroup$
Why D cannot be equal to 0?
$endgroup$
– Mathsaddict
Dec 3 '18 at 6:50
$begingroup$
Thank you for asking critical question. Please, see the updated answer.
$endgroup$
– farruhota
Dec 3 '18 at 15:55
$begingroup$
Thank you for asking critical question. Please, see the updated answer.
$endgroup$
– farruhota
Dec 3 '18 at 15:55
add a comment |
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You need only pick $a $ so that $f'(x)$ does not have two zeros. Use its discriminant.
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– coffeemath
Dec 3 '18 at 5:04
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Note: It's OK if $f'$ has only one zero. Then $f$ will still be strictly increasing or strictly decreaing, even though it is flat at a single point. [By strictly increasing is meant if $a<b$ then $f(a)<f(b).$ Similar for strictly decreasing.]
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– coffeemath
Dec 3 '18 at 5:53
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@coffeemath this is what i thought. Now, f is parabola, then it can be one one only if x takes positive real values. So, is it okay to take domain (0,infinity)?
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– Mathsaddict
Dec 3 '18 at 6:07
1
$begingroup$
No, look at formula for $f$ -- it's cubic so $f$ will not be a parabola. Its derivative will be. And if that derivative doesn't cross the axis twice, then the original $f$ will be 1-1 for all values of $x$ [i.e.domain all reals]
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– coffeemath
Dec 3 '18 at 6:20