Match the graphs of the functions in a–d with the graphs of th...

Question

Match the graphs of the functions in a–d with the graphs of their derivatives in A–D.

Accepted Answer

Below there. Today we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Given 4 graphs of functions labeled I through IV. And 4 graphs of their derivatives labeled A through D match the graph of each function with its corresponding derivative graph. Awesome. So it appears for this particular prompt, we're ultimately trying to figure out which of our original 4 functions match up to which of our 4 graphs for the derivatives. So we're trying to match the function graph with its derivative graph. That is what we're ultimately trying to do. So we're trying to figure out which derivative graph corresponds with which function graph. So, with that said, let's take a moment here to examine both sets of our graphs here. So, on our first coordinate plane, we have our functions I through IV, which are Roman numerals. So for our graph for I is a red parabola that is upright. And our graph for IV is a green parabola, which is upside down, pointing downwards, and then our graph for II is in black, and it appears to be a linear graph, or rather a linear curve, and our curve for III is in blue and it also appears to be linear in nature. Awesome, and for our 2nd coordinate plane, For our graphs A through D, we could see that we have a linear graph A that's in blue, and then all they all appear to be linear curves. So our curve in A is represented in blue, and it appears to be a horizontal line. Our curve for B, which is a diagonal line, is in green. And for C, we have another diagonal line which represents C and red, and all of these once again are linear curves. And finally, our final curve, D is a horizontal line denoted in black. Awesome. So it appears both curves B and C, both cross at the origin point. And it appears that the That are the curve for D. It's just a horizontal line just like A. OK. And it appears that it goes through. Y equals 1. Awesome. So with that said, now that we have a good visualization of what's happening in this particular problem, let us start to solve this problem officially. So first off, we need to note that for this particular problem, we can take two separate approaches to solve this problem. So for approach number 1, so for approach number 1, we need to recall and Note that the first derivative of a function corresponds to the slope of the tangent line to that specific function. So when we consider the functions I and IV, we can notice that each of them contains a horizontal tangent line at X equals 0. So, therefore, as a result, their derivative graphs should be equal to 0 at X equals 0. So it should be equal to 0. At X equals 0. So, this will mean that the graphs for I and IV will be matched with the derivative graphs B and C. So, Which means that that will leave I I and III to be matched with, so let's make a note of this, that I I. And I I I will be matched to, so I I I. And so I I and II, so our 2nd and 3rd graph will be matched with A and D. So now we need to consider the quadrants I and IV. So in quadrant I, if we draw a tangent line for function I at any point, the tangent line will have a positive slope. So this means that for X is greater than 0, the derivative graph should be positive. And this will be true for graph C, so that means that I is matched with C. Fantastic. So for quadrant IV, function IV has a negative slope of the tangent line at any point, and this corresponds to the line B that contains negative values in the whole quadrant. So therefore IV is matched with B. So now we need to consider function III, and we need to note and recall that it is a line with a positive slope, since it will increase with a value of X, so it's increasing with values of X. So there is also going to be an increase in Y, so such that that every point so at every point, a linear function will have the same slope. So that the derivative will be a constant function as a result. So moreover, that means that the constant is going to be positive, which allows us to match. Our function III with D. And for function II, we need to note that we can use a similar approach as we should recall, and we need to note that it will be a line with a negative slope with also an increase in the value of X as Y decreases. So therefore, as a result, the derivative would have to be a constant value that is negative, and we can match the graph Ii with A. Awesome. So now, for our second approach that we could take for this particular problem. is we can take a more advanced approach, which you may recall, and we need to be able to, in order to take this advanced approach, we need to be able to recall and recognize that I and IV are parabolas, and if we can identify them as parabolas, we can then differentiate, so using differentiation for their general forms, we would get that Y. is equal to AX2 + BX + C, and this will be all equal to 2 multiplied by a multiplied by X, or you can just simply write 2 A X to be. Precise, so is equal to 2 A X plus B, which is the equation of a line with a non-zero slope, and we need to note, like I mentioned before, that parabola I opens up, so that will mean That A is greater than 0. And this will yield a derivative that is a line with a positive slope which will be in graph C, or rather curve C. In contrast, IV opens downwards, so that will mean that A is less than 0, and that will mean that this will give us a negative slope for a line which will correspond to curve B. And for the remaining graphs, I I and III, those are linear functions in the form Y equals MX plus B, as I mentioned before, since they're linear looking in appearance or in nature, and this will mean that the derivative of this particular form will lead to Y, where Y is a fancy way of saying the first derivative of this function of Y. And this will be equal to MX plus B, which will be equal to M, which is a constant, and we can expect to get the derivative graphs that are that are going to be horizontal lines. So if a line has a positive slope where M is greater than 0, so that the corresponding derivative graph is above the X axis. And if the slope is negative, the graph of the derivative will be below the X axis. Awesome. So with that said, our final answers, without a doubt, and let's write them in green, that will mean for graph for curve I, it will be for C, and for II, it will be for A, and for III, it will be matched to D, and for IV it will be matched to B. And those are our final 4 answers. So to quickly recap, I will be matched to C, I I will be matched to A, III will be matched to D, and IV will be matched to B. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Match the graphs of the functions in a–d with the graphs of their derivatives in A–D. <MATCH A-D IMAGE>

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