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# Derivatives of Polynomial Functions - Problem 3

###### Norm Prokup

###### Norm Prokup

**Cornell University**

PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

To find points at which a derivative of a function has a certain value, first find the derivative. For a polynomial function, you can use the power rule and sum rule to find the derivative. Then, set the derivative equal to the value you are trying to find points for. From this, solve for x. The solutions will be the x-coordinates where this function has at value of derivative. To find the y-coordinates, simply evaluate the function at those x's (this is because y=f(x)).

Let's solve a tougher problem. Here we have the problem says the graph of y equals x³ minus 6x² minus 11x plus 8 has a slope equal to 4 at exactly two points. Find the coordinates of the points.

Now first of all, let me address this last sentence. 'Find the coordinates of the points' tells us that the problem wants the answer to be in the form of coordinate pair; x and Y coordinates. So we have to remember to find the y coordinates of the points as well. We'll find the x coordinates first, then the y.

Now what does it mean for the slope of this graph to equal 4? It means the derivative of this function is going to equal 4. So I need to find the derivative of that function dy/dx. So that's the derivative with respect to x of all this stuff; x³ minus 6x² minus 11x plus 8.

Now I'm going to combine, up till now I've been separating the steps of using a sum and cross multiple. I'm going to do it all in one step here. The derivative with respect to x of x³ minus 6 times the derivative with respect to x of x², plus the derivative with respect to x. I'm just going to take the minus 11x plus 8 by itself, because it's a linear function. I know its derivative is just going to be -11.

So the derivative of x³ is 3x². This is going to be minus 6 times the derivative of x² which 2x. So -6 times 2x minus 12x. Again this is -11. So what we want is, we want to find the x coordinates where this derivative equals 4. So let's set it equal to 4. This is going to give me the quadratic equation 3x² minus 12x, and I subtract 4 from both sides. I get minus 15 equals 0. So I have to solve this quadratic equation.

Now the first thing I'm noticing on the left side, is that all of these have 3 as a common factor. So I can divide that out, you don't lose any solutions by dividing out a constant. So dividing out the 3, I get x² minus 4x minus 5 equals 0. This looks factorable. So I'm going to see if I can factor it. I have an x and an x, and that will give me my x².

I need factors of -5 here. It looks like -5, and +1 is going to work here. -5 times +1 is going to give me -5. Equally importantly, I get -5x plus x is -4. So this works. This is my factored equation. This tells me that x equals 5, and x equals -1 are my solutions. Let me write that up here.

So these are the x coordinates of the two points where the slope is 4. All we have to do now is find the coordinates. To do that, we have to use this formula here. So I'm going to plug 5 into this formula. I get y equals 5³ which is 125, minus 6 times 5². 5² is 25. 6 times that is 150. So minus 150. Minus 11 times 5, that's minus 55, plus 8. So what do we have? -25 minus 55, is -80 plus 8, is -72. So one of my points is 5,-72.

Then for x equals -1. This one might be easier. When I plug -1 in I get -1³ is -1. -6 times -1² which is 1, so minus 6, minus 11 times -1, so +11, +8. We have -7 plus 8 is 1, plus 11 is 12. So our second point is -1,12. At these two 5,-72, and -1,12. The slope of the curve is 4.

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###### Norm Prokup

PhD. in Mathematics, University of Rhode Island

B.S. in Mechanical Engineering, Cornell University

He uses really creative examples for explaining tough concepts and illustrates them perfectly on the whiteboard. It's impossible to get lost during his lessons.

Thiswas EXCELLENT! I am a math teacher and have been looking for an easy/logical way to explain the lateral area of a cone to my students and this was incredibly helpful, thank you very much!”

I just learned more In 3 minutes of polygons here than I do in 3 weeks in my math class”

Hahaha, his examples are the same problems of my math HW!”

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