Mixing
Find and classify the stationary points of $y = x^ 2/(x-4)$
$begingroup$
I have already obtained the answer by using the quotient rule and so there are stationary points at x = 0 and x = 8. I am fine at doing these types of questions but I don't fully understand the meaning behind the answer. This was also provided in the solution of the answer
Now y approaches infinity as x approaches infinity, and
y approaches infinity as x approaches 4^(+). Hence, x = 8 must be a minimum. Similarly, y approaches (-)infinity as x approaches (-)infinity
and as x approaches 4^(-). Hence x = 0 is a maximum.
How do you know that x=8 is a maximum and x=0 is a minimum also what does the 4^(+) and 4^(-) mean ? sorry if this is a silly question I just don't get what the final answer says.
stationary-point
edited Jan 15 at 16:54
David G. Stork
11k41432
asked Jan 15 at 1:33
CinnaCinna
122
$endgroup$
add a comment |
$begingroup$
I have already obtained the answer by using the quotient rule and so there are stationary points at x = 0 and x = 8. I am fine at doing these types of questions but I don't fully understand the meaning behind the answer. This was also provided in the solution of the answer
Now y approaches infinity as x approaches infinity, and
y approaches infinity as x approaches 4^(+). Hence, x = 8 must be a minimum. Similarly, y approaches (-)infinity as x approaches (-)infinity
and as x approaches 4^(-). Hence x = 0 is a maximum.
How do you know that x=8 is a maximum and x=0 is a minimum also what does the 4^(+) and 4^(-) mean ? sorry if this is a silly question I just don't get what the final answer says.
stationary-point
edited Jan 15 at 16:54
David G. Stork
11k41432
asked Jan 15 at 1:33
CinnaCinna
122
$endgroup$
add a comment |
$begingroup$
I have already obtained the answer by using the quotient rule and so there are stationary points at x = 0 and x = 8. I am fine at doing these types of questions but I don't fully understand the meaning behind the answer. This was also provided in the solution of the answer
Now y approaches infinity as x approaches infinity, and
y approaches infinity as x approaches 4^(+). Hence, x = 8 must be a minimum. Similarly, y approaches (-)infinity as x approaches (-)infinity
and as x approaches 4^(-). Hence x = 0 is a maximum.
How do you know that x=8 is a maximum and x=0 is a minimum also what does the 4^(+) and 4^(-) mean ? sorry if this is a silly question I just don't get what the final answer says.
stationary-point
edited Jan 15 at 16:54
David G. Stork
11k41432
asked Jan 15 at 1:33
CinnaCinna
122
$endgroup$
I have already obtained the answer by using the quotient rule and so there are stationary points at x = 0 and x = 8. I am fine at doing these types of questions but I don't fully understand the meaning behind the answer. This was also provided in the solution of the answer
Now y approaches infinity as x approaches infinity, and
y approaches infinity as x approaches 4^(+). Hence, x = 8 must be a minimum. Similarly, y approaches (-)infinity as x approaches (-)infinity
and as x approaches 4^(-). Hence x = 0 is a maximum.
How do you know that x=8 is a maximum and x=0 is a minimum also what does the 4^(+) and 4^(-) mean ? sorry if this is a silly question I just don't get what the final answer says.
stationary-point
stationary-point
edited Jan 15 at 16:54
David G. Stork
11k41432
asked Jan 15 at 1:33
CinnaCinna
122
edited Jan 15 at 16:54
David G. Stork
11k41432
asked Jan 15 at 1:33
CinnaCinna
122
edited Jan 15 at 16:54
David G. Stork
11k41432
edited Jan 15 at 16:54
David G. Stork
11k41432
edited Jan 15 at 16:54
David G. Stork
11k41432
11k41432
asked Jan 15 at 1:33
CinnaCinna
122
asked Jan 15 at 1:33
CinnaCinna
122
asked Jan 15 at 1:33
CinnaCinna
122
122
add a comment |
add a comment |
1 Answer
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$begingroup$
Take the second derivatives to determine whether a function is maximized or minimized at a point.
$4^+$ means approaching the point $4$ from above.
answered Jan 15 at 1:44
David G. StorkDavid G. Stork
11k41432
$endgroup$
$begingroup$
if it is $y=frac{x^2}{x-4}$ it is a hyperbola with one vertical and one slanted asymptote. For some reason we've had several in the past few days
$endgroup$
– Will Jagy
Jan 15 at 2:32
$begingroup$
Okay thanks for the help I understand it now.
$endgroup$
– Cinna
Jan 16 at 2:07
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
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$begingroup$
Take the second derivatives to determine whether a function is maximized or minimized at a point.
$4^+$ means approaching the point $4$ from above.
answered Jan 15 at 1:44
David G. StorkDavid G. Stork
11k41432
$endgroup$
$begingroup$
if it is $y=frac{x^2}{x-4}$ it is a hyperbola with one vertical and one slanted asymptote. For some reason we've had several in the past few days
$endgroup$
– Will Jagy
Jan 15 at 2:32
$begingroup$
Okay thanks for the help I understand it now.
$endgroup$
– Cinna
Jan 16 at 2:07
add a comment |
$begingroup$
Take the second derivatives to determine whether a function is maximized or minimized at a point.
$4^+$ means approaching the point $4$ from above.
answered Jan 15 at 1:44
David G. StorkDavid G. Stork
11k41432
$endgroup$
$begingroup$
if it is $y=frac{x^2}{x-4}$ it is a hyperbola with one vertical and one slanted asymptote. For some reason we've had several in the past few days
$endgroup$
– Will Jagy
Jan 15 at 2:32
$begingroup$
Okay thanks for the help I understand it now.
$endgroup$
– Cinna
Jan 16 at 2:07
add a comment |
$begingroup$
Take the second derivatives to determine whether a function is maximized or minimized at a point.
$4^+$ means approaching the point $4$ from above.
answered Jan 15 at 1:44
David G. StorkDavid G. Stork
11k41432
$endgroup$
Take the second derivatives to determine whether a function is maximized or minimized at a point.
$4^+$ means approaching the point $4$ from above.
answered Jan 15 at 1:44
David G. StorkDavid G. Stork
11k41432
edited Jan 15 at 1:52
answered Jan 15 at 1:44
David G. StorkDavid G. Stork
11k41432
answered Jan 15 at 1:44
David G. StorkDavid G. Stork
11k41432
answered Jan 15 at 1:44
David G. StorkDavid G. Stork
11k41432
11k41432
$begingroup$
if it is $y=frac{x^2}{x-4}$ it is a hyperbola with one vertical and one slanted asymptote. For some reason we've had several in the past few days
$endgroup$
– Will Jagy
Jan 15 at 2:32
$begingroup$
Okay thanks for the help I understand it now.
$endgroup$
– Cinna
Jan 16 at 2:07
add a comment |
$begingroup$
if it is $y=frac{x^2}{x-4}$ it is a hyperbola with one vertical and one slanted asymptote. For some reason we've had several in the past few days
$endgroup$
– Will Jagy
Jan 15 at 2:32
$begingroup$
Okay thanks for the help I understand it now.
$endgroup$
– Cinna
Jan 16 at 2:07
$begingroup$
if it is $y=frac{x^2}{x-4}$ it is a hyperbola with one vertical and one slanted asymptote. For some reason we've had several in the past few days
$endgroup$
– Will Jagy
Jan 15 at 2:32
$begingroup$
if it is $y=frac{x^2}{x-4}$ it is a hyperbola with one vertical and one slanted asymptote. For some reason we've had several in the past few days
$endgroup$
– Will Jagy
Jan 15 at 2:32
$begingroup$
Okay thanks for the help I understand it now.
$endgroup$
– Cinna
Jan 16 at 2:07
$begingroup$
Okay thanks for the help I understand it now.
$endgroup$
– Cinna
Jan 16 at 2:07
add a comment |
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