Find all local maximum and minimum points of the function $f$. The 2019 Stack Overflow...
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Find all local maximum and minimum points of the function $f$.
The 2019 Stack Overflow Developer Survey Results Are InMaximum and Minimum ValueFinding all local maximum and minimum for some unusual functions.Finding intervals using local min and max (in interval notation form)Minimum/Maximum Question (Calculus)Local max, min and point of inflection of $y= ax^3 + 3bx^2 + 3cx - d$Find he local maximum and minimum value and saddle points of the function?Find the absolute minimum and maximum values of $f(theta) = cos theta$Maximum/MinimumDetermine local max., local min., and saddle points of the following function: $4x + 4y + x^2y + xy^2$$f$ differentiable $5$ times around $x=a, f'(a)=f''(a)=f'''(a)=0, f^{(4)}(x) <0 Rightarrow x=a$ is either a local minimum or a local maximum point
$begingroup$
Any help with this problem I have would be so much appreciated, i've been going round in circles and have no idea what I'm really doing.
I have the problem:
- Find all local maximum and minimum points of the function $f = xy$.
I've done the first derivative to get $$f' = 1(frac{dy}{dx})$$
But I have no clue on how to find the local max and min from this.
Any help would be grateful.
Thank you.
calculus maxima-minima
asked Mar 22 at 0:20
The StatisticianThe Statistician
117112
$endgroup$
add a comment |
$begingroup$
Any help with this problem I have would be so much appreciated, i've been going round in circles and have no idea what I'm really doing.
I have the problem:
- Find all local maximum and minimum points of the function $f = xy$.
I've done the first derivative to get $$f' = 1(frac{dy}{dx})$$
But I have no clue on how to find the local max and min from this.
Any help would be grateful.
Thank you.
calculus maxima-minima
asked Mar 22 at 0:20
The StatisticianThe Statistician
117112
$endgroup$
add a comment |
$begingroup$
Any help with this problem I have would be so much appreciated, i've been going round in circles and have no idea what I'm really doing.
I have the problem:
- Find all local maximum and minimum points of the function $f = xy$.
I've done the first derivative to get $$f' = 1(frac{dy}{dx})$$
But I have no clue on how to find the local max and min from this.
Any help would be grateful.
Thank you.
calculus maxima-minima
asked Mar 22 at 0:20
The StatisticianThe Statistician
117112
$endgroup$
Any help with this problem I have would be so much appreciated, i've been going round in circles and have no idea what I'm really doing.
I have the problem:
- Find all local maximum and minimum points of the function $f = xy$.
I've done the first derivative to get $$f' = 1(frac{dy}{dx})$$
But I have no clue on how to find the local max and min from this.
Any help would be grateful.
Thank you.
calculus maxima-minima
calculus maxima-minima
asked Mar 22 at 0:20
The StatisticianThe Statistician
117112
asked Mar 22 at 0:20
The StatisticianThe Statistician
117112
asked Mar 22 at 0:20
The StatisticianThe Statistician
117112
asked Mar 22 at 0:20
The StatisticianThe Statistician
117112
asked Mar 22 at 0:20
The StatisticianThe Statistician
117112
117112
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The local extrema of a function occur at stationary points, i.e. where the function's total derivative is $0$. In terms of partial derivatives, if the function has continuous partial derivatives (which this one does), this is equivalent to the partial derivatives both being $0$ at the point.
Partially differentiating, we have
begin{align*}
f_x &= y \
f_y &= x.
end{align*}
These are both $0$ only at $(x, y) = (0, 0)$.
However, this stationary point is neither a local minimum nor maximum. Consider, for $t in Bbb{R} setminus {0}$,
begin{align*}
f(t, t) &= t^2 > 0 = f(0, 0) \
f(t, -t) &= -t^2 < 0 = f(0, 0).
end{align*}
That is, there are points as close as you want to $(0, 0)$ that have greater and lesser function values than $f(0, 0) = 0$.
answered Mar 22 at 1:27
Theo BenditTheo Bendit
20.9k12355
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The local extrema of a function occur at stationary points, i.e. where the function's total derivative is $0$. In terms of partial derivatives, if the function has continuous partial derivatives (which this one does), this is equivalent to the partial derivatives both being $0$ at the point.
Partially differentiating, we have
begin{align*}
f_x &= y \
f_y &= x.
end{align*}
These are both $0$ only at $(x, y) = (0, 0)$.
However, this stationary point is neither a local minimum nor maximum. Consider, for $t in Bbb{R} setminus {0}$,
begin{align*}
f(t, t) &= t^2 > 0 = f(0, 0) \
f(t, -t) &= -t^2 < 0 = f(0, 0).
end{align*}
That is, there are points as close as you want to $(0, 0)$ that have greater and lesser function values than $f(0, 0) = 0$.
answered Mar 22 at 1:27
Theo BenditTheo Bendit
20.9k12355
$endgroup$
add a comment |
$begingroup$
The local extrema of a function occur at stationary points, i.e. where the function's total derivative is $0$. In terms of partial derivatives, if the function has continuous partial derivatives (which this one does), this is equivalent to the partial derivatives both being $0$ at the point.
Partially differentiating, we have
begin{align*}
f_x &= y \
f_y &= x.
end{align*}
These are both $0$ only at $(x, y) = (0, 0)$.
However, this stationary point is neither a local minimum nor maximum. Consider, for $t in Bbb{R} setminus {0}$,
begin{align*}
f(t, t) &= t^2 > 0 = f(0, 0) \
f(t, -t) &= -t^2 < 0 = f(0, 0).
end{align*}
That is, there are points as close as you want to $(0, 0)$ that have greater and lesser function values than $f(0, 0) = 0$.
answered Mar 22 at 1:27
Theo BenditTheo Bendit
20.9k12355
$endgroup$
add a comment |
$begingroup$
The local extrema of a function occur at stationary points, i.e. where the function's total derivative is $0$. In terms of partial derivatives, if the function has continuous partial derivatives (which this one does), this is equivalent to the partial derivatives both being $0$ at the point.
Partially differentiating, we have
begin{align*}
f_x &= y \
f_y &= x.
end{align*}
These are both $0$ only at $(x, y) = (0, 0)$.
However, this stationary point is neither a local minimum nor maximum. Consider, for $t in Bbb{R} setminus {0}$,
begin{align*}
f(t, t) &= t^2 > 0 = f(0, 0) \
f(t, -t) &= -t^2 < 0 = f(0, 0).
end{align*}
That is, there are points as close as you want to $(0, 0)$ that have greater and lesser function values than $f(0, 0) = 0$.
answered Mar 22 at 1:27
Theo BenditTheo Bendit
20.9k12355
$endgroup$
The local extrema of a function occur at stationary points, i.e. where the function's total derivative is $0$. In terms of partial derivatives, if the function has continuous partial derivatives (which this one does), this is equivalent to the partial derivatives both being $0$ at the point.
Partially differentiating, we have
begin{align*}
f_x &= y \
f_y &= x.
end{align*}
These are both $0$ only at $(x, y) = (0, 0)$.
However, this stationary point is neither a local minimum nor maximum. Consider, for $t in Bbb{R} setminus {0}$,
begin{align*}
f(t, t) &= t^2 > 0 = f(0, 0) \
f(t, -t) &= -t^2 < 0 = f(0, 0).
end{align*}
That is, there are points as close as you want to $(0, 0)$ that have greater and lesser function values than $f(0, 0) = 0$.
answered Mar 22 at 1:27
Theo BenditTheo Bendit
20.9k12355
answered Mar 22 at 1:27
Theo BenditTheo Bendit
20.9k12355
answered Mar 22 at 1:27
Theo BenditTheo Bendit
20.9k12355
answered Mar 22 at 1:27
Theo BenditTheo Bendit
20.9k12355
20.9k12355
add a comment |
add a comment |
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