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Question 45 f4 if, < } Compute Jo f ()d , glven f(x) = 0 9 ifr_ 2.621672 D 0 En15...

Question

Question 45 f4 if, < } Compute Jo f ()d , glven f(x) = 0 9 ifr_ 2.621672 D 0 En15

Question 4 5 f4 if, < } Compute Jo f ()d , glven f(x) = 0 9 ifr_ 2. 62 16 72 D 0 En15



Answers

Given each function, evaluate: $f(-1), f(0), f(2), f(4)$ $$ f(x)=\left\{\begin{array}{lll} 4 x-9 & \text { if } & x<0 \\ 4 x-18 & \text { if } & x \geq 0 \end{array}\right. $$

So in this problem, we have a piece wise function. These air parabolas these air What we're going to evaluate. So in this case, I'm starting at zero. In this case, when I plug zero in, I'm gonna get zero squared, which is zero. And I'm gonna get one. Since it's less than zero, I'm gonna have an open circle now. One. Now, if I were to plug in negative one negative one squared is a positive one. Positive. One plus one is two. If I plugged in negative too, I would end up with positive five. And so this is going to be a parabola. That looks like this. And then I could work on the second half again. If I plug in zero, I'm gonna get zero. Minus one is negative. One again. Open circle. Notice that at zero, there's no it's undefined because not neither are are filled ing. This one will end up going over one down, one over. Two down. 1234 Kind of the reciprocal. The opposite of one of the phone. And I'm gonna get a graph that looks something like this when I plug in. Negative to negative two is less than zero. So I'm going to use the first equation. Negative two squared plus one is four plus one, which is fine. There is my point. It matches perfect. Negative one is also less than zero. So negative one squared plus one is gonna give me too. As I look, it matches up when I put in positive. Wanted to use the second equation because my because positive one is greater than zero. So when I plug in the opposite of one squared minus one, so that's gonna end up giving me negative, too. And you'll notice it does match what the graph tells us it should be. And then positive to again. I'm gonna have the opposite of two squared minus one, which will end up giving me negative five, which matches up again to the graph.

This problem. We're working with a piece wise function. The first thing we have to do is sketch it. So let's work on the top one first when I plug in when x zero, I'm gonna have a value it negative, too. It's not going to equal that, but because it's not, there's no equal sign here. Since this is just a upside down parabola that shifted down to units, I'm going to go over one down, 1/2. 123 down four could make a table of values as well. Help me to get a picture for this one. This is an upward facing parabola at 02 will be an open circle of two because it's not equal to zero again over one up, one over to up or and we get an upward parabola here when we evaluate. Since negative two is less than zero, I'm going to use the first equation. Negative two Squared is positive for positive. For the opposite of that will give us negative for negative for minus two gives us negative six. Notice that to that end up with an output value negative six Negative one is also less than zero. The negative one squared is positive. One after that is negative. One negative one minus two is negative. Three knows it matches the graph. One happens to be greater than zero, so I can use the second equation. One squared is one plus two is three notice. 13 is the point that matches up on the craft. Two is greater than zero. So get him using a second equation. So I'm going to be plugging in. Two squared, which is four plus two is six. Notice that to six we it matches up to the graph. So here's my evaluation. Here's my graph. This is a piece of ice function.

Okay. We want to find where the derivative is positive and where the derivative is negative. So let's find the derivative is a product. So I'll use the product rule first times the derivative of the second plus the second times the derivative of the first -8 x plus 24 -8 X -14. So -16 x plus 10. All right. Set it equal to zero. So minus 16 X equals minus 10 X equals 26th or five eights. Okay, Because we're gonna find where it's equal to zero because that's where it's gonna change signs. If it does put it on a number line and then put the derivative, Pick a number bigger than five. -16 times 100 plus 10. That's negative. Pick a number less than 5.80 negative 16 times zero plus 10. That's positive. So F prime of X is negative for X between 5/8 and infinity and F prime of X is positive for numbers from minus infinity up to five eights.

We have. Let's see here first. Let me fix this because I messed that up again. One and two. So we have um Alright, function is two X squared plus X. To the two thirds power minus four Differentiating this we get four x. Um plus two thirds X. To the minus one third power. We're told there's a zero between one and two and we're gonna use C1 equals 1.5. That's our initial cost, Plug in one, plug in to we get different -1 in about 5.5. So we can see that there should be since this is continuous and that reigns there should be a zero in their between here and here. Now this is probably not a great initial guess because I would assume that it's probably closer to one than two, 10: 1.5. Close to the one than to to sow. I would've probably said let's just .1.25 is my initial best looking at these things. So the function at this value here is 1.8, I'm sorry, he said um First iteration of Newton's method, we kind of getting 1.22 497. So again we are moving down as expected From looking at this so we get our function value of 0.1459 second iteration, we get 1.1985 so on and a function value when obviously we have to um Yeah two Zeros after the decimal. Um So it isn't always not not great yet, but we're still moving down. We see so forth, innovation, we get 1.1 982959 So on. And now we get you know the one time 7 -7 for the value of function. So that's pretty good. And you can see that we have these digits now that seem to be accurate and then we go to 1/5 duration and we can see that these digits seem to be accurate right? So and in fact now we get it up with one times 10 to the minus 15th as our value for the function, so this is really close um to the zero.


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