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the curve $v= 0.00005(t-200)^2 - 1$ seems to have only $1$ minimum point to me at $(200,-1)$ as a minimum point. (i completed the squares to find it) but in my book it says it also has a maximum point at $(0,1)$

$0\leq t\leq800$

my question is how would you have found the maximum point? i learned the trigonometric graphs of sine and cosine and know that they have multiple turning points but how can a quadratic graph have multiple turning points? could anyone please explain why?

sweetie sakura
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  • Welcome to MSE. A local extrema is a zero of the derivative. And a linear only has at most one real zero. – Countable Feb 04 '21 at 16:55
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    If the restriction is $0 \le t \le 800$ then the endpoints of this restriction can be maxes. – coffeemath Feb 04 '21 at 16:55
  • yes but how would you have calculated them? please answer.i really need help. – sweetie sakura Feb 04 '21 at 16:56
  • Welcome to MSE. Please use MathJax to format your posts. To begin with, surround math expressions (including numbers) with $ signs and use _ for subscripts. $x_1$ comes out as $x_1$. – saulspatz Feb 04 '21 at 16:57
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    Does your book say anything about $(800,17)$, or does it only mention $(0,1)$? – Barry Cipra Feb 04 '21 at 17:03
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    sakura-- Just plug each endpoint into the function and compare results to the value of the function at the critical point $t=200.$ – coffeemath Feb 04 '21 at 17:04
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    @sweetie sakura. You ask "yes but how would you have calculated them?" You know your function ($v$) begins at $t=0$ and $t=800$ So all you have to do is calculate the value of $v$ for these values. And then compare with other possible minimums and maximums. – MasB Feb 04 '21 at 17:05
  • no it does not mention (800, 17). and coffeemath i still dont understand the logic behind what you're saying. could you please elaborate? – sweetie sakura Feb 04 '21 at 17:05
  • so what you're saying @BernardMassé is that the extremities can be turning points along with the turning point that is obtained with completing the squares. but please, explain your logic. – sweetie sakura Feb 04 '21 at 17:08
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    A global maximum needn't be a "turning point". It's simply that value of $t\in[0,800]$ for which the function gets its highest value. – Intelligenti pauca Feb 04 '21 at 17:08

1 Answers1

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When looking for a curve's maxima or minima, you need to examine two things - inflection points and discontinuities. Inflection points are the turning points that you seem to be familiar with, which can be found through the typical use of derivatives. Derivatives don't exist at discontinuous points, however, so you need to check them individually. This parabolic function is discontinuous at the endpoints of t=0 and t=800, so you need to check values there. Basically, you can have maxima/minima at points where the slope is not equal to zero - these aren't "turning points", but rather endpoints or some other kind of discontinuity. You can see, for example, that the straight line x=y defined for [0,10] has no inflection points at all, but still has a maximum and minimum.

Nuclear Hoagie
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  • How is the function "discontinuous" at the endpoints? – Robert Israel Feb 04 '21 at 17:45
  • @RobertIsrael Hm, you might be right that that's not an appropriate choice of words. What I'm getting at is that there is is only a left- or right-sided limit at the endpoints, although you don't really need the other one since approaching from the other side is not in the domain of the function. We don't have the usual case where left- and right-sided limits are equal in a continuous function, although I'm not sure that's really sufficient for discontinuity, depending on the definition. – Nuclear Hoagie Feb 04 '21 at 17:57
  • I would simply say "you need to examine three things: stationary points, singular points, and endpoints". Stationary points are points in the interior of the interval where the derivative is $0$. Singular points are points in the interior of the interval where the derivative does not exist. – Robert Israel Feb 04 '21 at 21:31
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    By the way, inflection points are not what you seem to think they are. They are points where the curvature changes, thus the second derivative (if it exists) is $0$. – Robert Israel Feb 04 '21 at 21:33