Finding stationary points (1 Viewer)

[davidbarnes]

"Find any stationary points on the curve f(x) = X^4 - 2X^2 - 3."
I can always find the first point, although I don't know how you are supposed to find the second (or in some cases third point) that the answer shows.

Likewise with "Show that the curve f(x) = X^5 + 1 has a point of inflexion at (0, 1)." - I get that it is a local minimum for some reason and can't seem to work out how to show the SP is inflexion.

Finally with one such as "A curve has f'(x) = X(x+1), for what x values does the curve have stationary points?" - This one looked quite basic at first, although in the end I don't really have a clue how to go about it.

Any help woudl be greatly appreciated.

1st problem - I can find 1 SP, but not 2.
2nd problem - Can't solve.
3rd problem - Not sure how to do, book has no examples.

 

[watatank]

"Find any stationary points on the curve f(x) = X^4 - 2X^2 - 3."
I can always find the first point, although I don't know how you are supposed to find the second (or in some cases third point) that the answer shows.

Likewise with "Show that the curve f(x) = X^5 + 1 has a point of inflexion at (0, 1)." - I get that it is a local minimum for some reason and can't seem to work out how to show the SP is inflexion.

Finally with one such as "A curve has f'(x) = X(x+1), for what x values does the curve have stationary points?" - This one looked quite basic at first, although in the end I don't really have a clue how to go about it.

Any help woudl be greatly appreciated.

1st problem - I can find 1 SP, but not 2.
2nd problem - Can't solve.
3rd problem - Not sure how to do, book has no examples.

first problem:

f(x) = X^4 - 2X^2 - 3.
f'(x) = 4x^3 - 4x = 4x(x^2 - 1) = 4x(x+1)(x-1)

stat points are the values of x where dy/dx = 0. as you can see there's three stationary points, where x = 0, +/- 1. sub the respective x points into the original function and you get your y coordinate

second: hmm thought i could do it, seems as though i found it being a local minimum as well

third: they've given you the gradient function, make that equal zero and solve for x. so its just 0, -1.

 

[ssglain]

Second:
f(x) = x^5 + 1
f'(x) = 5x^4
f"(x) = 20x^3

For this one, you need to check the concavity on both sides of the stationary point (0, 1) in order to determine its nature.
<TABLE style="WIDTH: 192pt; BORDER-COLLAPSE: collapse" cellSpacing=0 cellPadding=0 width=256 border=0 x:str><COLGROUP><COL style="WIDTH: 48pt" span=4 width=64><TBODY><TR style="HEIGHT: 12.75pt" height=17><TD class=xl22 style="BORDER-RIGHT: #ece9d8; BORDER-TOP: #ece9d8; BORDER-LEFT: #ece9d8; WIDTH: 48pt; BORDER-BOTTOM: #ece9d8; HEIGHT: 12.75pt; BACKGROUND-COLOR: transparent" width=64 height=17>x</TD><TD class=xl22 style="BORDER-RIGHT: #ece9d8; BORDER-TOP: #ece9d8; BORDER-LEFT: #ece9d8; WIDTH: 48pt; BORDER-BOTTOM: #ece9d8; BACKGROUND-COLOR: transparent" width=64>0-</TD><TD class=xl22 style="BORDER-RIGHT: #ece9d8; BORDER-TOP: #ece9d8; BORDER-LEFT: #ece9d8; WIDTH: 48pt; BORDER-BOTTOM: #ece9d8; BACKGROUND-COLOR: transparent" width=64 x:num>0</TD><TD class=xl22 style="BORDER-RIGHT: #ece9d8; BORDER-TOP: #ece9d8; BORDER-LEFT: #ece9d8; WIDTH: 48pt; BORDER-BOTTOM: #ece9d8; BACKGROUND-COLOR: transparent" width=64>0+</TD></TR><TR style="HEIGHT: 12.75pt" height=17><TD class=xl22 style="BORDER-RIGHT: #ece9d8; BORDER-TOP: #ece9d8; BORDER-LEFT: #ece9d8; BORDER-BOTTOM: #ece9d8; HEIGHT: 12.75pt; BACKGROUND-COLOR: transparent" height=17>f"(x)</TD><TD class=xl22 style="BORDER-RIGHT: #ece9d8; BORDER-TOP: #ece9d8; BORDER-LEFT: #ece9d8; BORDER-BOTTOM: #ece9d8; BACKGROUND-COLOR: transparent">negative</TD><TD class=xl22 style="BORDER-RIGHT: #ece9d8; BORDER-TOP: #ece9d8; BORDER-LEFT: #ece9d8; BORDER-BOTTOM: #ece9d8; BACKGROUND-COLOR: transparent" x:num>0</TD><TD class=xl22 style="BORDER-RIGHT: #ece9d8; BORDER-TOP: #ece9d8; BORDER-LEFT: #ece9d8; BORDER-BOTTOM: #ece9d8; BACKGROUND-COLOR: transparent">positive</TD></TR></TBODY></TABLE>The change in sign of f"(x) indicates a change in concavity, which means that (1, 0) a point of horizontal inflexion.

Remember to always translate the words into their mathematical meaning. Take the third one, stationary point means that f'(x) = 0, as watatank said.

 

[Grace HSC]

for the secong one about the points of inflexion, if u sub the x value (in this case 0) into the second derivative of f(x) you get zero, and when f''(x) is equal to 0 there is usually a point of inflexion, then check ur concav on either side by subbing an x value on either side of the SP into the second derivative. :wave: