Posted by andy on Sunday, July 22, 2012 at 1:41pm.
from what you have done correctly, I made the following sketch
Graph rises out of quadr III to a max point (-3,175) in II , turns around at that point and has a point of inflection at (-2.12,110), has another point of inflection at (0,10), and continued to drop to (3,-152)in quadrant IV for a minimum point. It then rises and continues to do so into quadrant I
Did you notice that at (0,10)both y' and y'' = 0, thus you have a point inflection and a turning point.
When that happens there will be a point where the graph "levels off".
In general, at a point of inflection, the graph will have the shape of the letter S.
So from there, the graph is concave down from -infinity to -2.12, concave up from -2.12 to 0, concave down from 0 to 2.12, and finally concave up from 2.12 to +infinity
I often use this online grapher to look at the functions
http://rechneronline.de/function-graphs/
enter: x^5 - 15x^3 + 10 in "first graph" window
change setting in "Range y-axis from" to -300 and +300
you might also want to click on "derivative" after studying the original graph.
or to make it even more interesting, enter the first derivative in "second graph"
and the second derivative in"third graph"
WOW! Thanks Much! Love that site you gave me and I totally understand and can plainly see it all except the max and min. I am seeing the max at (-3, 175) and the min at (3, -152) but does that satisfy "if x is restricted to the interval {1, 5)" because I thought that meant between +1 and +5 on the x-axis? And did I get the correct "critical points", (0,10), 3,-152),-3, 175)?Critical point cannot be the same as max and min can they?
ahh, I didn't notice the part about 1 ≤ x ≤ 5
- that means you want to consider the graph of the function only in that domain,
f(1) = 1^5 - 15(1^3) = -14
f(5) = 5^5 - 15(5^3) = 3125 - 1875 = 1250
let's change the "Range x-axis from" to 1 to 5 and
change the "Range y-axis from" from -300 to 1300
We can see that the minimum value of y is -152 when x=3 and the maximum value is 1250 when x=5
Usually the word "critical points" refer to either maximum or minimum points, or to points where the derivative does not exist.
Some texts refer to critical points as any point where something "important" happens.
So I am ttinking the critical points are the (-3,175)(3, -152)
So I am correct in my thinking the critical points are the (-3,175)(3 -152) because that was asked of the original function,.... because I only have to give only the max and min values for y if x is restricted to the interval {1,5}? Thanks again for all your help.
not for the given domain of x from 1 to 5, since x=-3 lies outside that domain.
What was the actual wording of the question ?
the original function was x^5-15x^3 +10 so when you figured the min and max with the restrictions and substituted 1, and 5 for the x's should you have included the constant 10? So I came up with (-4) and (1260), but forgive me I still don't know what that means...thes are the y values for the max and min when x is restricted to intervals {1, 2} ?
The question was 6-fold but the only part with the restriction was min and max values of y...everything else pertained to the original function.Thanks for staying with me...it's been a long day!
Of course I should have included the +10 in the calculation, I really don't know why I didn't
(you get those temporary "mind gaps" when you get my age, ha ha)
f(1) = 1^5 - 15(1^3) + 10 = -4
etc
When you look at the graph between x=1 and x=5,
what is the lowest value of y you get or see ? y = -4
what is the largest value of y you see ? y = 1260
That's really all there is to this.
That rechneronline software is crisp and full of options!
I'll definite keep that in the toolbox.
Thanks Reiny!
Yes, it is one of the best online graphing tools I found.
From the .de extension it looks like it comes out of Germany.
I especially like the fact that you can overlay 3 graphs, very useful to show transformations.
I use it often in conjunction with Wolfram's pages.
Of course the hundreds of clips on Khan Academy are great as well.
Well thanks again for the help...and I am probably older than you...hence the mind gaps of my own!