1. (a) Sketch the graph of y = ln x. (b) On that graph...

1. (a) Sketch the graph of $y = \ln x$. (b) On that graph color the area that corresponds to the value of $\int_0^1 \ln x \, dx$. Note: that we say \"corresponds to\", not \"equals to\". If a function is negative, the value of the integral is equal to the negative of the area that we can mark on a graph. (c) The function $y = \ln x$ is the inverse of $y = e^x$. Recall that the graph of an inverse function is obtained from the graph of $y = f(x)$ by reflection about the line $y = x$. On the same graph sketch the graph of $y = e^x$. (d) Color the reflection about the line $y = x$ of the area marked in part (b). (e) Based on the picture, compute the value of $\int_0^1 \ln x \, dx$ as an integral of function $e^x$ (with the proper limits!).

Transcript

00:01 Ok, so i have put up a screen grab of the graph here, and hopefully this is similar to the graph that's alluded to in your question.

00:11 It's the exact same function that's been written, which is f of x is x cubed minus 3x minus 2.

00:20 Ok, so for the first question, we want to show calculation of the function values for x equals minus 3 and x equals 3.

00:28 So we're looking for f of minus 3.

00:31 So that's going to be minus 3 cubed minus 3 times minus 3 minus 2.

00:39 So minus 3 cubed is minus 27, and 3 times minus 3 is minus 9, so we have a plus 9, and then minus 2.

00:48 So minus 27 plus 9 minus 2 gives us minus 20.

00:53 So that's f of minus 3, and f of 3 is going to be 3 cubed minus 3 times 3 minus 2, so that's 27 minus 9 minus 2.

01:06 So 27 minus 9 minus 2 gives us 16.

01:11 So that's what we have for 3.

01:15 Now you can check that that makes sense.

01:17 If you look down at x equals 3, and when you reach the graph at x equals minus 3, make sure that the y value is minus 20.

01:26 And likewise at x equals 3, meet the graph and make sure that the x value is 16, and we have that in both cases.

01:34 Ok, and now we want to read off from the graph the function values when x is minus 2.

01:39 So that gives us a y value of minus 4.

01:45 So when x is minus 2, we get y is minus 4.

01:52 And then when x is 0, we can see that y is going to be minus 2.

01:59 And then when x is 1, again, here's x is 1, look down on the y, onto the graph, see where we end up, and we end up at y is minus 4.

02:15 Ok, so now we want to check that the function can also be written in its factorized form.

02:22 So what we can do is just expand these brackets.

02:27 So x plus 1 squared is x plus 1 times x plus 1, and then it's x minus 2.

02:32 So i'm going to expand the first two brackets first.

02:35 We get x squared plus x plus x, so that's plus 2x plus 1, and that's multiplied by x minus 2.

02:43 And then we just need to expand these brackets here.

02:45 So we get x cubed minus 2x squared plus 2x squared minus 4x plus x minus 2, and that gives us x cubed minus 3x minus 2, and that's what we wanted.

03:04 Ok, so use this to calculate the zero points of f.

03:08 So if we use this to calculate the zero points of f, that means that f is going to be 0.

03:12 So x plus 1 squared times x minus 2 equals 0, and that can only happen if the first bracket equals 0 or the second bracket equals 0.

03:24 So if the first bracket is 0, then x must be minus 1, and that's what we call a repeated root because it happens twice because of the squared.

03:33 And if the second bracket is 0, x is 2.

03:37 So see if that makes sense on the graph.

03:39 We see that the graph comes up and just touches at minus 1, so that's the repeated root, and then at x equals 2 it goes straight through the axis.

03:49 Ok, so for b we want to find the derivative, we want to find f prime of x.

03:56 So that's going to be 3x squared minus 3.

04:01 And if we solve this for it being equal to 0, that becomes x squared equals 3 over 3, or x squared equals 1, and that has a solution where x is plus or minus 1.

04:16 Ok, and let's see that that makes sense.

04:18 Is the derivative 0 at x equals minus 1 and x equals plus 1? yes, absolutely it is.

04:24 The derivative is 0 right there.

04:26 Ok, now we can read from the graph where the function is increasing.

04:30 Well, we can see it's increasing when x is less than minus 1 and when x is greater than 1.

04:36 So we have that the graph is increasing.

04:40 There's a couple of different ways to write this, i'm going to write it like this.

04:43 When x is less than minus 1 and when x is greater than 1.

04:49 In fact, i'll change that and to an or, since they can't be true at the same time.

04:53 So x is less than minus 1 or x is greater than 1.

04:57 And where was it decreasing? it was just decreasing in between those two points.

05:03 So when minus 1 is less than x is less than 1.

05:07 Ok, and now we want to find the local extrema.

05:11 So all we want to do is work out whether x equals plus 1 is a maxima or a minima.

05:18 And you can do this by finding the second derivative if you don't want to just look at the graph.

05:22 So the second derivative is 6x, where i've just differentiated 3x squared to get 6x...

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