Taking the derivative of f(x)=x³-12x+2 and graphing the derivative, so we can tell when f is increasing or decreasing, and where is its relative extremum points. Created by Sal Khan.
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- How can we figure out the maxima or minima without graphing the function? Is there a way?(91 votes)
- Take the derivative of the function and find where that equals 0 to find critical points. Then take the second derivative and find its value at the critical points. If the second derivative is positive, then the point is a minimum; if it's negative, then it's a maximum; if it's zero, then it's an inflection point.
Sal talks about it here: https://www.khanacademy.org/math/calculus/derivative_applications/critical_points_graphing/v/concavity--concave-upwards-and-concave-downwards-intervals(144 votes)
- How do we know that the graph will pass over to the positive side of the graph it could also remain in the negative side by changing its concavity?(9 votes)
- Good question, and it is related to something you will be studying soon.
To recap: f(x)=x^3-12x+2 and f'(x)=3x^2-12
We found two critical points of f(x) via f'(x), x=-2 and x=2, so we know the function will pass from the negative side of the y axis (x<0), to the positive side (x>0).
When x=-2, we have f(-2) = (-2)^3 - 12(-2) + 2 = -8 - (-24) + 2 = -8 + 24 + 2 = 18
When x=2, we have f(2) = (-)^3 - 12(2) + 2 = 8 - (24) + 2 = 8 - 24 + 2 = -14
So we know as x moves from -2 to 2, f(x) will move from y=18 to y=-14.
And this is how we know the graph will pass over to the positive side of both the x and y axes.
Had there been anymore concavity changing points (there is only 1 at x=0), they would have been hinted at via the first derivative test, so you can be sure that the graph is concave down for all x<0 and concave up for all x>0. What you will learn soon are inflection points, where concavity changes, and how the are found, via the 2nd derivative test. In this case, given that the first derivative is f'(x)=3x^2-12, the second derivative is f''(x)=6x, and it is only zero at x=0, so x=0 is the only place where the graph changes concavity.
You might want to try this great tool that graphs function to help you get an intuition of the relationship between the degree of a function and its behavior. https://www.desmos.com/calculator(14 votes)
- how did you find the derivative 3x^2 -12(4 votes)
- Taking the derivative is actually very easy:
First what you have to do is to write down the original function f(x), which would be f(x)=x^3-12x-5 in this case!
now you look at all x-variables...
take the power of the x-variable and multiply it by the coefficient and subtract the power by one; like this: 1*x^3 -> 1*3x^3-1 -> 3x^2 !
next step: 12x = 12x^1 -> 12*1x^1-1 -> 12x^0 [anything raised to zero equals 1] -> 12*1 -> 12
last step: "5".. there is no variable, so this you just can cross out
result: f'(x)=3x^2-12(11 votes)
- Can i just substitute 2 and -2 in the original function and conclude that the greater f(x) value as a result of the substitution is the maximum point and the lesser is the minimum point?(7 votes)
- In second derivative test , if the value is less than 0 why is it maximum ?? Also if greater than 0 why is it minimum??(6 votes)
- If the value of the second derivative is less than 0 (a negative value) then it means the shape of the original function is concave down (resembling a hill or bump or upside down "u", or a parabola opening downward) which means the critical point in the interval of the concave down shape will be the highest point in the interval (the peak of the hill) and thus it is a local maximum.
If the value of the second derivative is more than 0 (a positive value) then the shape of the original function is concave up (resembling a bowl, a dent/pit, or a "u" shape, or a parabola opening upwards) and the critical value in a concave up shape is the lowest point in the interval at the very base/bottom of the "pit" and thus it is a local minimum.(3 votes)
- So there are no other local minima or maxima points in that function or are these global maxima and minima points?(2 votes)
- as far as I have seen in the lecture Sal has drawn the graph of f'(x) rather than f(x) and according to my understanding The relative extrema's and absolute extrema's are part of f(x) graph itself rather than its derivative. Hope you get the idea(1 vote)
- I need help trying to solve f(x)= 2x^3+3x^2-72x. I have a problem seting the critical value to 0 and solving.(1 vote)
- First factor out x
f(x) = x*(2x^2+3x-72)
To find the roots, set f(x) = 0 and solve for x values.
One will be x = 0, and the other two will come from the quadratic formula applied to 2x^2+3x-72.(6 votes)
- Find the area of the largest isosceles triangle that can be inscribed in a circle of radius 4
a)solve by writing the area as a function of h
b)solve by writing the area s a function of a
c) identify the type of triangle of maximum area
Im really aggravated with this problem because I did part of part a, which is to write the function... A=1/2(2x)94+h). I cant solve it because Im getting a different derivative compared to others.
And as for b) I know we have to use trig but I dont know how it implement it with regard to area.(2 votes)
- I'm having trouble finding the second derivative for (x+1)/ sqrt(x), so that I can look for points of concavity
I'm getting a weird value(2 votes)
- The second derivative is:
(3-x)/ (4 √x⁵)
Here's how I'd recommend doing this derivative:
y =(x+1)/ √(x)
y = x/√x + 1/√x
y = x^(½) + x^(-½)
dy/dx = ½x^(-½) - ½ x^(-³⁄₂)
d²y/dx² = -¼ x^(-³⁄₂) + ¾x^(-⁵⁄₂)
Factor ¼ x^(-⁵⁄₂)
= ¼ x^(-⁵⁄₂) [ -x + 3]
= (3-x) / (4 √x⁵)(1 vote)
- To figure out whether the critical point is a maximum or minimum, can I simply plug in a value for x ?
Ex: In this video, the derivative of f(x) is 3x^2 - 12. We know that the derivative is zero or that the critical points are at 2 and -2. To find if critical point is a maximum or minimum for critical point 2. Why can't I just do f ( 1 ) = 3 ( 1 ) ^ 2 - 12 which equals a negative number so I know that the function is negative before the critical point from the Left. No I plug in f ( 3 ) = 3 ( 3 ) ^ 2 - 12, I get a positive number so I know the function is positive after passing the critical point. So I have figured out that the critical point at 2 is a minimum because before 2 the slope is negative and after 2 the slope is positive. IS this an accurate way of finding out whether or not the critical point is a maximum or minimum? Is there a better, more valid way of doing this?
Thank you(2 votes)
- Yes, that works. Whether what you described (which is the First Derivative Test) is the easiest way depends on whether your original function or your derivative is easier to do the math with. So, once you have the critical points, just decide whether it is easier to use f(x) or f'(x) to do the further computations.
What you left out is that if f'(x) has the same sign on both sides of the critical point, then you don't have an extremum at all, but instead you have a saddle point.
Yet another way to find whether a critical point is a max or a min is the second derivative test -- frequently easier, but it does not always work. Here's how to do it:
If f'(c) = 0 (this method doesn't work for critical points where f'(c) does not exist),
If f''(c) > 0 then f(c) is a minimum ← Note these are second derivatives, not first
If f''(c) < 0 then f(c) is a maximum
If f''(c) = 0 or if f''(c) does not exist, then you cannot tell by this method(1 vote)
We've got the function f of x is equal to x to the third power minus 12x plus 2. And what I want to do in this video is think about at what points does my function f take on minimum or maximum values? And to figure that out I have to first figure out what are the critical points for my function f. And then which of those critical points do we achieve a minimum or maximum value? And to determine the critical points we have to find the derivative of our function because our critical points are just the point at which our derivative is either equal to 0 or undefined. So the derivative of this thing right over here, we're just going to use the power rule several times, and then I guess you can call it the constant rule. But the derivative of x to the third is 3x squared. Derivative of negative 12x is negative 12. And the derivative of a constant, it doesn't change with respect to x, so it's just going to be equal to 0. So we're going to get a critical point when this thing right over here, for some value of x is either undefined or 0. Well this thing is defined for all values of x. So the only places we're going to find critical points is when this thing is equal to 0. So let's set it equal to 0. When does 3x squared minus 12 equal 0? So let's add 12 to both sides. You get 3x squared is equal to 12. Divide both sides by 3. You get x squared is equal to 4. Well this is going to happen when x is equal to 2 and x is equal to negative 2. Just to be clear, f of 2, or let me be clear, f prime of 2, you get 3 times 4 minus 12, which is equal to 0. And f prime negative 2 is also, same exact reason, is also equal to 0. So we can say-- and I'll switch colors here-- that f has critical points at x equals 2 and x equals negative 2. Well that's fair enough. But we still don't know whether the function takes on a minimum values at those points, maximum values of those points, or neither. To figure that out we have to figure out whether the derivative changes signs around these points. So let's actually try to graph the derivative to think about this. So let's graph. I'll draw an axis right over here. I'll do it down here because maybe we can use that information later on to graph f of x. So let's say this is my x-axis. This is my y-axis. And so we have critical points at x is equal to positive 2. So it's 1, 2. And x is equal to negative 2, 1, 2. x is equal to negative 2. So what does this derivative look like if we wanted to graph it? Well we have when x is equal to 0 for the derivative we're at negative 12. So this is the point y is equal to negative 12. So this is, we're graphing y is equal to f prime of x. So it looks something like this. These are obviously the 0's of our derivative. So it has to move up to cross the x-axis there and over here. So what is the derivative doing at each of these critical points? Well over here our derivative is crossing from being positive, we have a positive derivative, to being a negative derivative. So we're crossing from being a positive derivative to being a negative derivative, that was our criteria for a critical point to be a maximum point. Over here we're crossing from a negative derivative to a positive derivative, which is our criteria for a critical point for the function to have a minimum value at a critical point. So a minimum. And I just want to make sure we have the correct intuition. If our function, if some function is increasing going into some point, and at that point we actually have a derivative 0-- the derivative could also be undefined-- but we have a derivative of 0 and then the function begins decreasing, that's why this would be a maximum point. Similarly, if we have a situation where the function is decreasing going into a point, the derivative is negative. Remember this is the graph of the derivative. Let me make this clear. This is the graph of y is equal to not f of x, but f prime of x. So if we have a situation we're going into the point, the function has a negative slope, we see we have a negative slope here-- so the function might look something like this. And then right at this point the function is either undefined or has 0 slope. So in this case it has 0 slope. And then after that point, let me do it right under it. So going into it we have a negative slope. And then right over here we have a 0 slope. Which I could draw it even better than that. So if we were to imagine going into it we have a negative slope, right at that point we have a 0 slope, and then we have a positive slope. So the function begins increasing. That's why we say we have a minimum point right over there. So what I did it right over here is to try to conceptualize what the function itself could look like given the derivative, in this case switching from a positive derivative to a negative derivative, across that critical point, or going from a negative derivative to a positive derivative. That's why this is the criteria for a maximum point, this is the criteria for a minimum point. Well with that out of the way, can we use this intuition that we just talked about to at least try to sketch the graph of f of x? So let's try to do it. And it's just going to be a sketch, it's not going to be very exact. But at least it'll give us a sense of the shape of what f of x looks like. So my best attempt. So it might not be drawn completely to scale. So it's my x-axis, this is my y-axis. We know we have a critical point at x is equal to positive 2. And we have a critical point at x is equal to negative 2. We know just from inspection that the y-intercept right here, if the graph of y is equal to f of x, when x is 0 f of x is 2. So we're going to hit right over-- I don't want to draw this completely to the same scale as the x-axis. So let's say that this is 2 right over here. So this is where we're going to cross. This is going to be our y-intercept. And so we said already that we have a maximum point at x is equal to negative 2. So what is f of negative 2? f of negative 2 is equal to 8 or negative 8, let me be careful. It's negative 8. And then we're going to have 12 times negative 2, which is negative 24. But then we're going to add it. So we're subtracting negative 24. So this is plus 24. And then we finally have plus 2. So negative 8 plus 24 plus 2, so that's going to be negative, let's see, negative 8 plus 24 is 16, plus 2 is 18. So f of negative 2 is equal to 18. And I'm not drawing it completely to scale. But let's say that this is 18 right over here. So this is the function. This is the point negative 2 comma 18. And we know that it's a maximum point. The derivative going into that point is negative. The derivative going into that point is negative-- sorry, the derivative going into that point is positive. So we are increasing. The slope is positive. And then after we cross that point, the slope becomes negative. The derivative cross the x-axis, the slope becomes negative. I actually want to use that same color. It looks like this. And then, of course, the graph is going to cross the-- it's going to have a y-intercept, something like that. And then, as we approach 2, we are approaching another critical point. Now what is f of 2? f of 2 is going to be equal to positive 8 minus 24 plus 2. So this is 10 minus 24, which is equal to negative 14. So let's say this is the point negative 14 right over here. Actually I can draw it a little bit. Let's say this is negative 14. So this is f of 2 right over there. And we saw already that the slope is negative as we approach it. So our function is decreasing as we approach it. And then right there the slope is 0. We figured that out earlier, that's how we identified it being a critical point. And then the slope is increasing after that, the derivative is positive. The slope is increasing. So this is our sketch of what f of x could look like given that these are the critical points. And we were able to identify 2 as a minimum point. So this was a minimum value. The function takes on a minimum value when x is equal to 2. And the function took on a maximum value when x was equal to negative 2.