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For what value of $h$ is the minima equal to $-5$
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I have some questions regarding this exercise.
$f_h(x)=x^4-hx^2$
"For what value of $h$ is the minima equal to $-5$"
What I would have thought is correct:
$f_h'(x)=(4x^3-2hx)=0$
$=x(4x^2-2h)=0$
$x=0$ and $x=pmsqrtfrach2$
Then, I would say okay, well if $h=10$ then this would satisfy the question. Meaning the answer is $h=10$. But for some reason, the answer sheet states the following steps instead:
Take $x=pmsqrtfrach2$ and input into the function:
$f(-sqrtfrach2)=f(sqrtfrach2)$
=
$frach^22=5$ Therefore, $h=sqrt10$.
I don't quite get how that way worked. They figured out the slope, put it into the original equation and equaled it to $-5$, how does that give them the required value.
Would appreciate some explanation, thanks in advance.
calculus algebra-precalculus derivatives
asked Jul 26 at 22:13
user472288
390111
add a comment |Â
up vote
0
down vote
favorite
I have some questions regarding this exercise.
$f_h(x)=x^4-hx^2$
"For what value of $h$ is the minima equal to $-5$"
What I would have thought is correct:
$f_h'(x)=(4x^3-2hx)=0$
$=x(4x^2-2h)=0$
$x=0$ and $x=pmsqrtfrach2$
Then, I would say okay, well if $h=10$ then this would satisfy the question. Meaning the answer is $h=10$. But for some reason, the answer sheet states the following steps instead:
Take $x=pmsqrtfrach2$ and input into the function:
$f(-sqrtfrach2)=f(sqrtfrach2)$
=
$frach^22=5$ Therefore, $h=sqrt10$.
I don't quite get how that way worked. They figured out the slope, put it into the original equation and equaled it to $-5$, how does that give them the required value.
Would appreciate some explanation, thanks in advance.
calculus algebra-precalculus derivatives
asked Jul 26 at 22:13
user472288
390111
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have some questions regarding this exercise.
$f_h(x)=x^4-hx^2$
"For what value of $h$ is the minima equal to $-5$"
What I would have thought is correct:
$f_h'(x)=(4x^3-2hx)=0$
$=x(4x^2-2h)=0$
$x=0$ and $x=pmsqrtfrach2$
Then, I would say okay, well if $h=10$ then this would satisfy the question. Meaning the answer is $h=10$. But for some reason, the answer sheet states the following steps instead:
Take $x=pmsqrtfrach2$ and input into the function:
$f(-sqrtfrach2)=f(sqrtfrach2)$
=
$frach^22=5$ Therefore, $h=sqrt10$.
I don't quite get how that way worked. They figured out the slope, put it into the original equation and equaled it to $-5$, how does that give them the required value.
Would appreciate some explanation, thanks in advance.
calculus algebra-precalculus derivatives
asked Jul 26 at 22:13
user472288
390111
I have some questions regarding this exercise.
$f_h(x)=x^4-hx^2$
"For what value of $h$ is the minima equal to $-5$"
What I would have thought is correct:
$f_h'(x)=(4x^3-2hx)=0$
$=x(4x^2-2h)=0$
$x=0$ and $x=pmsqrtfrach2$
Then, I would say okay, well if $h=10$ then this would satisfy the question. Meaning the answer is $h=10$. But for some reason, the answer sheet states the following steps instead:
Take $x=pmsqrtfrach2$ and input into the function:
$f(-sqrtfrach2)=f(sqrtfrach2)$
=
$frach^22=5$ Therefore, $h=sqrt10$.
I don't quite get how that way worked. They figured out the slope, put it into the original equation and equaled it to $-5$, how does that give them the required value.
Would appreciate some explanation, thanks in advance.
calculus algebra-precalculus derivatives
asked Jul 26 at 22:13
user472288
390111
asked Jul 26 at 22:13
user472288
390111
asked Jul 26 at 22:13
user472288
390111
asked Jul 26 at 22:13
user472288
390111
390111
add a comment |Â
add a comment |Â
3 Answers
3
active
oldest
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up vote
1
down vote
accepted
$x=0$ and $x=pmsqrtfrac h2$
Then, I would say okay, well if $h=10$ then this would satisfy the question.
What? Why? I don't understand this at all.
....
I don't quite get how that way worked. They figured out the slope,
Not quite, this isn't a line so there is no single slope. They figured out what the derivative was and this gives a formula for the slope (of the tangent line) at any given $x$.
What the actually figured out was the three points where the slope will be equal to $0$. At $x = 0$ then the slope of the tangent line is $0$. ANd $x= sqrtfrac h2$ the slope of the tangent line is $0$ and at $x= -sqrtfrac h2$ the slope of the tangent line is $0$. At all other values of $x$ the slope of the tangent line is not $0$.
put it into the original equation
Not quite. The slope at $x_0$ is $0$. They did not put the slope into the original equation. They put the value of $x$ where $f'(x) = 0$ back into the original equation.
So when $f'(x) = 0$ then $f(x)$ is at an extreme value.
and equaled it to −5.
Because we were told $f(x) = -5$ when $x$ is such that $f(x)$ is a minimum value.
how does that give them the required value.
It's a basic observation, that extrema points (max, mins, and saddles) of continuous functions can only and always appear when the slope of the tangent line is $0$.
So .... if $(x_0, f(x_0))$ is a local minimum, then $f'(x_0) = 0$. And if $f'(x_0) =0$ then $x_0 = 0$ or $x_0 =pm sqrtfrac h2$.
So either $(0, f(0)) = (0, 0)$ is the minimum value $(0, -5)$ (which is impossible as $0 ne -5$.
or $(sqrtfrac h2, f(sqrtfrac h2)) = (sqrtfrac h2, frac h^24 - h* frac h2)= (sqrtfrac h2, -frac h^24)$ is the minimum value of $(sqrtfrac h2, -5)$
Or $(-sqrtfrac h2, f(-sqrtfrac h2) = (-sqrtfrac h2, frac h^24 - h* frac h2)= (-sqrtfrac h2, -frac h^24)$ is the minimum value of $(-sqrtfrac h2, -5)$
As $f(0) = 0 > -5$ that is neither the minimum nor is it $-5$.
So $f(pm sqrtfrac h2 = -frac h^24 = -5$ are the two local minima.
So what is the $h$ that will make this true?
So we solve for $-frac h^24 = -5$.
Which is $h=pm sqrt20$. But for $x = pm sqrt frac h2$ to be defined we must have $hge 0$ so $h = sqrt20$.
answered Jul 26 at 23:19
fleablood
60.3k22575
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up vote
1
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The minimum of the function should be $-5$ . It means if $f'(a) = 0$ then $f(a) = -5$ . Here we have $a = pm sqrt h/2$
answered Jul 26 at 22:22
S.H.W
1,0431720
add a comment |Â
up vote
1
down vote
you have combined things that should be written separately. For a chosen $h,$ the minimum value occurs when $x = pm sqrt h/2.$ Let's see, the original function was $f_h(x)=x^4 - h x^2.$ Alright
$$ f_h( sqrt h/2 ) = frach^24 - frach^22 = - frach^24 ; . $$
They want this to be $-5,$ so we need $$ - frach^24 = -5, $$
$$ h^2 = 20, $$
but $h$ must be positive so
$$ h = sqrt20 = 2 sqrt 5$$
Always wise to check. We have $frach2 = sqrt 5,$ so that the $x$ value we are using is $5^1/4.$ Then the function value becomes
$$ 5 - 2 sqrt 5 sqrt 5 = 5 - 10 = -5 $$
Note that $sqrt20 approx 4.47 ; , ;$ while $5^1/4 approx 1.49534878$ is pretty close to $frac32$
answered Jul 26 at 22:24
Will Jagy
96.9k594195
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
$x=0$ and $x=pmsqrtfrac h2$
Then, I would say okay, well if $h=10$ then this would satisfy the question.
What? Why? I don't understand this at all.
....
I don't quite get how that way worked. They figured out the slope,
Not quite, this isn't a line so there is no single slope. They figured out what the derivative was and this gives a formula for the slope (of the tangent line) at any given $x$.
What the actually figured out was the three points where the slope will be equal to $0$. At $x = 0$ then the slope of the tangent line is $0$. ANd $x= sqrtfrac h2$ the slope of the tangent line is $0$ and at $x= -sqrtfrac h2$ the slope of the tangent line is $0$. At all other values of $x$ the slope of the tangent line is not $0$.
put it into the original equation
Not quite. The slope at $x_0$ is $0$. They did not put the slope into the original equation. They put the value of $x$ where $f'(x) = 0$ back into the original equation.
So when $f'(x) = 0$ then $f(x)$ is at an extreme value.
and equaled it to −5.
Because we were told $f(x) = -5$ when $x$ is such that $f(x)$ is a minimum value.
how does that give them the required value.
It's a basic observation, that extrema points (max, mins, and saddles) of continuous functions can only and always appear when the slope of the tangent line is $0$.
So .... if $(x_0, f(x_0))$ is a local minimum, then $f'(x_0) = 0$. And if $f'(x_0) =0$ then $x_0 = 0$ or $x_0 =pm sqrtfrac h2$.
So either $(0, f(0)) = (0, 0)$ is the minimum value $(0, -5)$ (which is impossible as $0 ne -5$.
or $(sqrtfrac h2, f(sqrtfrac h2)) = (sqrtfrac h2, frac h^24 - h* frac h2)= (sqrtfrac h2, -frac h^24)$ is the minimum value of $(sqrtfrac h2, -5)$
Or $(-sqrtfrac h2, f(-sqrtfrac h2) = (-sqrtfrac h2, frac h^24 - h* frac h2)= (-sqrtfrac h2, -frac h^24)$ is the minimum value of $(-sqrtfrac h2, -5)$
As $f(0) = 0 > -5$ that is neither the minimum nor is it $-5$.
So $f(pm sqrtfrac h2 = -frac h^24 = -5$ are the two local minima.
So what is the $h$ that will make this true?
So we solve for $-frac h^24 = -5$.
Which is $h=pm sqrt20$. But for $x = pm sqrt frac h2$ to be defined we must have $hge 0$ so $h = sqrt20$.
answered Jul 26 at 23:19
fleablood
60.3k22575
add a comment |Â
up vote
1
down vote
accepted
$x=0$ and $x=pmsqrtfrac h2$
Then, I would say okay, well if $h=10$ then this would satisfy the question.
What? Why? I don't understand this at all.
....
I don't quite get how that way worked. They figured out the slope,
Not quite, this isn't a line so there is no single slope. They figured out what the derivative was and this gives a formula for the slope (of the tangent line) at any given $x$.
What the actually figured out was the three points where the slope will be equal to $0$. At $x = 0$ then the slope of the tangent line is $0$. ANd $x= sqrtfrac h2$ the slope of the tangent line is $0$ and at $x= -sqrtfrac h2$ the slope of the tangent line is $0$. At all other values of $x$ the slope of the tangent line is not $0$.
put it into the original equation
Not quite. The slope at $x_0$ is $0$. They did not put the slope into the original equation. They put the value of $x$ where $f'(x) = 0$ back into the original equation.
So when $f'(x) = 0$ then $f(x)$ is at an extreme value.
and equaled it to −5.
Because we were told $f(x) = -5$ when $x$ is such that $f(x)$ is a minimum value.
how does that give them the required value.
It's a basic observation, that extrema points (max, mins, and saddles) of continuous functions can only and always appear when the slope of the tangent line is $0$.
So .... if $(x_0, f(x_0))$ is a local minimum, then $f'(x_0) = 0$. And if $f'(x_0) =0$ then $x_0 = 0$ or $x_0 =pm sqrtfrac h2$.
So either $(0, f(0)) = (0, 0)$ is the minimum value $(0, -5)$ (which is impossible as $0 ne -5$.
or $(sqrtfrac h2, f(sqrtfrac h2)) = (sqrtfrac h2, frac h^24 - h* frac h2)= (sqrtfrac h2, -frac h^24)$ is the minimum value of $(sqrtfrac h2, -5)$
Or $(-sqrtfrac h2, f(-sqrtfrac h2) = (-sqrtfrac h2, frac h^24 - h* frac h2)= (-sqrtfrac h2, -frac h^24)$ is the minimum value of $(-sqrtfrac h2, -5)$
As $f(0) = 0 > -5$ that is neither the minimum nor is it $-5$.
So $f(pm sqrtfrac h2 = -frac h^24 = -5$ are the two local minima.
So what is the $h$ that will make this true?
So we solve for $-frac h^24 = -5$.
Which is $h=pm sqrt20$. But for $x = pm sqrt frac h2$ to be defined we must have $hge 0$ so $h = sqrt20$.
answered Jul 26 at 23:19
fleablood
60.3k22575
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
$x=0$ and $x=pmsqrtfrac h2$
Then, I would say okay, well if $h=10$ then this would satisfy the question.
What? Why? I don't understand this at all.
....
I don't quite get how that way worked. They figured out the slope,
Not quite, this isn't a line so there is no single slope. They figured out what the derivative was and this gives a formula for the slope (of the tangent line) at any given $x$.
What the actually figured out was the three points where the slope will be equal to $0$. At $x = 0$ then the slope of the tangent line is $0$. ANd $x= sqrtfrac h2$ the slope of the tangent line is $0$ and at $x= -sqrtfrac h2$ the slope of the tangent line is $0$. At all other values of $x$ the slope of the tangent line is not $0$.
put it into the original equation
Not quite. The slope at $x_0$ is $0$. They did not put the slope into the original equation. They put the value of $x$ where $f'(x) = 0$ back into the original equation.
So when $f'(x) = 0$ then $f(x)$ is at an extreme value.
and equaled it to −5.
Because we were told $f(x) = -5$ when $x$ is such that $f(x)$ is a minimum value.
how does that give them the required value.
It's a basic observation, that extrema points (max, mins, and saddles) of continuous functions can only and always appear when the slope of the tangent line is $0$.
So .... if $(x_0, f(x_0))$ is a local minimum, then $f'(x_0) = 0$. And if $f'(x_0) =0$ then $x_0 = 0$ or $x_0 =pm sqrtfrac h2$.
So either $(0, f(0)) = (0, 0)$ is the minimum value $(0, -5)$ (which is impossible as $0 ne -5$.
or $(sqrtfrac h2, f(sqrtfrac h2)) = (sqrtfrac h2, frac h^24 - h* frac h2)= (sqrtfrac h2, -frac h^24)$ is the minimum value of $(sqrtfrac h2, -5)$
Or $(-sqrtfrac h2, f(-sqrtfrac h2) = (-sqrtfrac h2, frac h^24 - h* frac h2)= (-sqrtfrac h2, -frac h^24)$ is the minimum value of $(-sqrtfrac h2, -5)$
As $f(0) = 0 > -5$ that is neither the minimum nor is it $-5$.
So $f(pm sqrtfrac h2 = -frac h^24 = -5$ are the two local minima.
So what is the $h$ that will make this true?
So we solve for $-frac h^24 = -5$.
Which is $h=pm sqrt20$. But for $x = pm sqrt frac h2$ to be defined we must have $hge 0$ so $h = sqrt20$.
answered Jul 26 at 23:19
fleablood
60.3k22575
$x=0$ and $x=pmsqrtfrac h2$
Then, I would say okay, well if $h=10$ then this would satisfy the question.
What? Why? I don't understand this at all.
....
I don't quite get how that way worked. They figured out the slope,
Not quite, this isn't a line so there is no single slope. They figured out what the derivative was and this gives a formula for the slope (of the tangent line) at any given $x$.
What the actually figured out was the three points where the slope will be equal to $0$. At $x = 0$ then the slope of the tangent line is $0$. ANd $x= sqrtfrac h2$ the slope of the tangent line is $0$ and at $x= -sqrtfrac h2$ the slope of the tangent line is $0$. At all other values of $x$ the slope of the tangent line is not $0$.
put it into the original equation
Not quite. The slope at $x_0$ is $0$. They did not put the slope into the original equation. They put the value of $x$ where $f'(x) = 0$ back into the original equation.
So when $f'(x) = 0$ then $f(x)$ is at an extreme value.
and equaled it to −5.
Because we were told $f(x) = -5$ when $x$ is such that $f(x)$ is a minimum value.
how does that give them the required value.
It's a basic observation, that extrema points (max, mins, and saddles) of continuous functions can only and always appear when the slope of the tangent line is $0$.
So .... if $(x_0, f(x_0))$ is a local minimum, then $f'(x_0) = 0$. And if $f'(x_0) =0$ then $x_0 = 0$ or $x_0 =pm sqrtfrac h2$.
So either $(0, f(0)) = (0, 0)$ is the minimum value $(0, -5)$ (which is impossible as $0 ne -5$.
or $(sqrtfrac h2, f(sqrtfrac h2)) = (sqrtfrac h2, frac h^24 - h* frac h2)= (sqrtfrac h2, -frac h^24)$ is the minimum value of $(sqrtfrac h2, -5)$
Or $(-sqrtfrac h2, f(-sqrtfrac h2) = (-sqrtfrac h2, frac h^24 - h* frac h2)= (-sqrtfrac h2, -frac h^24)$ is the minimum value of $(-sqrtfrac h2, -5)$
As $f(0) = 0 > -5$ that is neither the minimum nor is it $-5$.
So $f(pm sqrtfrac h2 = -frac h^24 = -5$ are the two local minima.
So what is the $h$ that will make this true?
So we solve for $-frac h^24 = -5$.
Which is $h=pm sqrt20$. But for $x = pm sqrt frac h2$ to be defined we must have $hge 0$ so $h = sqrt20$.
answered Jul 26 at 23:19
fleablood
60.3k22575
answered Jul 26 at 23:19
fleablood
60.3k22575
answered Jul 26 at 23:19
fleablood
60.3k22575
answered Jul 26 at 23:19
fleablood
60.3k22575
60.3k22575
add a comment |Â
add a comment |Â
up vote
1
down vote
The minimum of the function should be $-5$ . It means if $f'(a) = 0$ then $f(a) = -5$ . Here we have $a = pm sqrt h/2$
answered Jul 26 at 22:22
S.H.W
1,0431720
add a comment |Â
up vote
1
down vote
The minimum of the function should be $-5$ . It means if $f'(a) = 0$ then $f(a) = -5$ . Here we have $a = pm sqrt h/2$
answered Jul 26 at 22:22
S.H.W
1,0431720
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The minimum of the function should be $-5$ . It means if $f'(a) = 0$ then $f(a) = -5$ . Here we have $a = pm sqrt h/2$
answered Jul 26 at 22:22
S.H.W
1,0431720
The minimum of the function should be $-5$ . It means if $f'(a) = 0$ then $f(a) = -5$ . Here we have $a = pm sqrt h/2$
answered Jul 26 at 22:22
S.H.W
1,0431720
answered Jul 26 at 22:22
S.H.W
1,0431720
answered Jul 26 at 22:22
S.H.W
1,0431720
answered Jul 26 at 22:22
S.H.W
1,0431720
1,0431720
add a comment |Â
add a comment |Â
up vote
1
down vote
you have combined things that should be written separately. For a chosen $h,$ the minimum value occurs when $x = pm sqrt h/2.$ Let's see, the original function was $f_h(x)=x^4 - h x^2.$ Alright
$$ f_h( sqrt h/2 ) = frach^24 - frach^22 = - frach^24 ; . $$
They want this to be $-5,$ so we need $$ - frach^24 = -5, $$
$$ h^2 = 20, $$
but $h$ must be positive so
$$ h = sqrt20 = 2 sqrt 5$$
Always wise to check. We have $frach2 = sqrt 5,$ so that the $x$ value we are using is $5^1/4.$ Then the function value becomes
$$ 5 - 2 sqrt 5 sqrt 5 = 5 - 10 = -5 $$
Note that $sqrt20 approx 4.47 ; , ;$ while $5^1/4 approx 1.49534878$ is pretty close to $frac32$
answered Jul 26 at 22:24
Will Jagy
96.9k594195
add a comment |Â
up vote
1
down vote
you have combined things that should be written separately. For a chosen $h,$ the minimum value occurs when $x = pm sqrt h/2.$ Let's see, the original function was $f_h(x)=x^4 - h x^2.$ Alright
$$ f_h( sqrt h/2 ) = frach^24 - frach^22 = - frach^24 ; . $$
They want this to be $-5,$ so we need $$ - frach^24 = -5, $$
$$ h^2 = 20, $$
but $h$ must be positive so
$$ h = sqrt20 = 2 sqrt 5$$
Always wise to check. We have $frach2 = sqrt 5,$ so that the $x$ value we are using is $5^1/4.$ Then the function value becomes
$$ 5 - 2 sqrt 5 sqrt 5 = 5 - 10 = -5 $$
Note that $sqrt20 approx 4.47 ; , ;$ while $5^1/4 approx 1.49534878$ is pretty close to $frac32$
answered Jul 26 at 22:24
Will Jagy
96.9k594195
add a comment |Â
up vote
1
down vote
up vote
1
down vote
you have combined things that should be written separately. For a chosen $h,$ the minimum value occurs when $x = pm sqrt h/2.$ Let's see, the original function was $f_h(x)=x^4 - h x^2.$ Alright
$$ f_h( sqrt h/2 ) = frach^24 - frach^22 = - frach^24 ; . $$
They want this to be $-5,$ so we need $$ - frach^24 = -5, $$
$$ h^2 = 20, $$
but $h$ must be positive so
$$ h = sqrt20 = 2 sqrt 5$$
Always wise to check. We have $frach2 = sqrt 5,$ so that the $x$ value we are using is $5^1/4.$ Then the function value becomes
$$ 5 - 2 sqrt 5 sqrt 5 = 5 - 10 = -5 $$
Note that $sqrt20 approx 4.47 ; , ;$ while $5^1/4 approx 1.49534878$ is pretty close to $frac32$
answered Jul 26 at 22:24
Will Jagy
96.9k594195
you have combined things that should be written separately. For a chosen $h,$ the minimum value occurs when $x = pm sqrt h/2.$ Let's see, the original function was $f_h(x)=x^4 - h x^2.$ Alright
$$ f_h( sqrt h/2 ) = frach^24 - frach^22 = - frach^24 ; . $$
They want this to be $-5,$ so we need $$ - frach^24 = -5, $$
$$ h^2 = 20, $$
but $h$ must be positive so
$$ h = sqrt20 = 2 sqrt 5$$
Always wise to check. We have $frach2 = sqrt 5,$ so that the $x$ value we are using is $5^1/4.$ Then the function value becomes
$$ 5 - 2 sqrt 5 sqrt 5 = 5 - 10 = -5 $$
Note that $sqrt20 approx 4.47 ; , ;$ while $5^1/4 approx 1.49534878$ is pretty close to $frac32$
answered Jul 26 at 22:24
Will Jagy
96.9k594195
edited Jul 26 at 22:49
answered Jul 26 at 22:24
Will Jagy
96.9k594195
answered Jul 26 at 22:24
Will Jagy
96.9k594195
answered Jul 26 at 22:24
Will Jagy
96.9k594195
96.9k594195
add a comment |Â
add a comment |Â
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