Stationary Points: Exercise 1

by Batool Akmal

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    00:01 So for the next three exercise questions, we’ve been asked to find the stationary points and the nature.

    00:06 I hope you enjoyed going through them and actually sketched them for yourselves and saw how amazing it is to actually know what these graphs look like graphically and what they look like algebraically. But let’s try it out and check your solutions. So our first exercise question says we have y = x² + 1. Recall back to what happens when you’re looking for a stationary point.

    00:32 What is your gradient? If you remember correctly, at stationary points, dy/dx = 0. Dy/dx means differentiate so we just differentiate this function. When we differentiate this, we’ll get dy/dx = 2x.

    00:53 Remember to equal it to zero because you’re looking for stationary points. Generally, the gradient is 2x.

    00:59 Only at the stationary point does the gradient equal to zero. Nice and straightforward calculation, if 2x = 0, then x = 0. So that’s your coordinate of x. You can find your y which is important when you’re sketching it. So we want to know where exactly it is on the XY coordinate system.

    01:20 We can rewrite this as y = 0² + 1. So x² + 1, that gives me 1. Therefore the coordinates of your stationary points are (0, 1). Quickly doing a little sketch on the side to show us where everything is, (0, 1) x = 0, y = 1. So we’re here. Our stationary point is at (0, 1). Let’s call that y = 1. Now, we need to find out what the graph is actually doing. So what kind of stationary point is this? In order to do that, we’re going to do the second differential. Dy/dx = 2x. To find the nature, we’re going to do our second differential. D²y/dx² = 2. These are the easiest kind where you don’t have to substitute any x values.

    02:09 You’ve just got a numerical value as it is. This is greater than zero. Greater than zero means that it’s positive which makes it a minimum point. So this graph here would look like a minimum.

    02:21 You can just extend it to make it look like your x² graph. Now for the ones who are familiar with this graph, when you look at this expression here, hopefully you are thinking of what does an x² graph looks like anyway. It’s a positive parabola going through (0, 0). So we did that in one of the examples.

    02:40 The +1 here has just moved it up by 1 unit. So we could’ve predicted what this looks like anyway.

    02:47 But we’ve proved it and we’ve gone through the solution to actually show it accurately on our graph.

    About the Lecture

    The lecture Stationary Points: Exercise 1 by Batool Akmal is from the course Stationary Points.

    Included Quiz Questions

    1. (3, - 9)
    2. (6, 0)
    3. ( 0, 0)
    4. (-3, 27)
    5. (-6, 72)
    1. Maximum at (0, 2)
    2. Minimum at (0, 2)
    3. Point of inflection at (1/3, 5/3)
    4. Minimum at (1/3, 5/3)
    5. Maximum at (2/3, 2/3)

    Author of lecture Stationary Points: Exercise 1

     Batool Akmal

    Batool Akmal

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