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LT1CJILProblem 1. Suppose that f(5) = 1, f'(5) = 6, 9(5) = -3, and 9(5) = 2. Find the following values:(a) (fg)(5) (b) (f9)(5) (c) (g/f)(5)Problem 2. Find th...

Question

LT1CJILProblem 1. Suppose that f(5) = 1, f'(5) = 6, 9(5) = -3, and 9(5) = 2. Find the following values:(a) (fg)(5) (b) (f9)(5) (c) (g/f)(5)Problem 2. Find the derivative ofWJ"f(c)

LT1CJIL Problem 1. Suppose that f(5) = 1, f'(5) = 6, 9(5) = -3, and 9(5) = 2. Find the following values: (a) (fg)(5) (b) (f9)(5) (c) (g/f)(5) Problem 2. Find the derivative of WJ" f(c)



Answers

For the functions in Problems: (a) Find the derivative at $x=-1$ (b) Find the second derivative at $x=-1$ (c) Use your answers to parts (a) and (b) to match the function to one of the graphs in Figure $3.9,$ each of which is shown centered on the point (-1,-1) . $$f(x)=2 x^{3}+3 x^{2}-2$$

For this problem. In part they were given the function F of x Y equals x Y squared plus X squared. And we are asked to find the directional derivative of that function at the point negative +12 in the direction of U equals one half I plus three over to J. So our first step is to find the gradient of our function at X Y. So that is going to be let's see here. It's going to be y squared plus two, X I Plus two XYJ had that means that when we evaluate this at the point negative 12 then it will have two squared minus two. So we'll have just two I. Hat then plus two times negative one at times too. So minus four J hat. Now to find the directional derivative then we will have to take the dot product of that with the vector U where U is already normalized. So we'll have one half times two in the eye hat. So that will just be one then plus a route 3/2 times negative four. So it'll be minus two. Route three. With the result being approximately -2.46. For part B. Were asked to interpret the results as a rate of change. I'm going to be saying this verbally Well just right listen for part B and C. So for part B, what this is saying is that starting at the point negative 12 and moving in the direction of you. The function is changing at a rate of approximately negative 2.46 units per unit length, and then for part C. What this tells us is that the slope of the tangent line to the curve that we can consider as the intersection of the surface, F x y equals x squared plus X squared and the plane perpendicular to the xy plane that contains the point negative +12 and negative three. That would be the result, evaluating at that point In the direction of you is approximately -2.46.

So we have this function H. Of X. And in part a of our question, we're going to find the derivative or each prime of X at negative one. And so the way that we're gonna do this is we're gonna use the power rule which I will show right now, just just when we take the derivative of any X rays to some power, this is equal to that power times X rays to the N -1 or 1- Our Original Power. So knowing this, we're going to be able to actually find our derivative H. Prime of X. Which is equal to two times four, which is eight X to the four minus one. Which would be three Plus eight times 3 is 24 X squared 15 times two is 30 X. And then 14 times one is 14 X to the zero, which is just 14 times one. So and for constants like this for here all constants go to zero and they take the derivative. So this H prime of X is our derivative. So now we can just plug in -1. So h prime of negative one is equal to eight times negative one of the third is negative eight Plus 24 times negative one squared is 24 -30 plus 14. And this is actually equal to zero. So R H prime of negative one is equal to zero. And now for part B All we're gonna do is find the second derivative of our function at -1. And how we do this is we find the derivative of our H. Prime of X. And we're going to use the power rule again. So H double prime of X is equal to three times eight which is 24 X squared plus 24 times two X. Which is 48 X plus 30 times one X. The zeroth which is just 30. And now we can just plug in negative one again. So if we do this we have H double prime of negative one, it's equal to 24 times negative one squared is 20 for minus 48 plus 30. And this actually comes out to six so H double prime of X or of negative one is equal to six. And now for part C we're going to use the fact that H prime of negative one is equal to zero and H double prime of negative one is equal to six to actually match our function to a graph. So this means that we are neither increasing nor decreasing at this point. We're actually changing from increasing to decreasing at this point. So knowing that along with the fact that our H double prime of negative one being greater than zero means that we are concave up at And X is equal to negative. one means that the only graph from our options actually matches these two conditions is graph one. So the graph that we choose to match our function is graph one.

We are given the linear function F of X is defined as three X plus. Um the Square Root of two. Um So there's a short answer and then there's a longer explanation. I guess I can give you both. You may have heard me say that this is a linear problem, because if you were to look at these values, uh you know, the y intercept is the square root of two, which is about 1.41 And then the slope is up three, right one. Uh So if you think about how the derivative is finding the slope at a point, well, what's the slope at all of these points? They're all equal to three. So that is probably the short explanation. But if you would actually rewrite this out and I might help some of you by telling you the square to two is about 1.414. If I asked you for the derivative of three X. I hope you would all say, okay, well, it's just three. Again, it's a linear piece and then the slope of 1.414 The slope of a constant is always equal to zero. Well, we don't write plus zero because three plus zero is just three. So that's your correct answer three.

So we're given this G fx function. And for part a of our question, we're going to try and find the derivative of our function at negative one. And so how we're gonna do this is we're going to find the derivative of G. Of X. Using the power rule. So G prime of X is equal to if you know the power rule, what you do is you take this exponents appear, you bring it down so we have four X. And then in the exponent now we just minus one. So we have four minus one and our exponents. So this is X to the third. And we can do this with all of our exes that are raised to some exponents. So this minus two minus X squared becomes minus two X to the first and then this minus two, X becomes minus two X to the zero or just minus two. And our constant negative three goes to zero. All constants. When we're taking derivatives go to zero. So now that we have this derivative of our function, we can just plug in negative one to find G. Prime of negative one. So four times negative one to the third is negative four minus two. X times negative one plus two minus two. So this is equal to negative four plus two is negative two minus two is negative four. And now for part B we're gonna find the second derivative of our function at negative one. So the first thing we're gonna do is find the second derivative of our function. And how we do that is we just look at our first derivative and then find the derivative of that. So we're gonna use the power rule again here since we have a four X to the third, we're gonna bring the three down. So we have four times three which is 12 X. To the three minus one, which would be two X to the second. And then for minus two X. We bring the X. We bring the X. Of the first. So we bring the one down so we have minus two times X. To the zero or just minus two. And now that we have this second derivative of our function we can just plug a negative one again. So negative one, negative one squared is one. So we just have 12 minus two. So this is equal to 10. And now for part C we're gonna use the fact that G prime of -1 was equal to negative four. And we're going to use the the fact that our g double prime or a second derivative at negative one was equal to 10. And because our first derivative was equal to negative four at negative one, we know that we are decreasing at at the point negative one on our function G. Of X. And since we have a positive value for our second derivative at negative one, we know that we are concave up at this point. So if we ever have a second derivative that is positive at a point, that means that we are concave up at that point. So because we know these two things, the only graph that were given in our options that actually matches these two is graph number two. So that is our choice for the graph that matches our function.


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