Sprint3.13 PM7 $ 239mathweb sandbox csun eduHomework 4: Problem 2PreviousProblem ListNextpoint) Find the value of the constant that makes the following function con...

You may use a graphing calculator to solve the following problems.
If $f$ has a derivative at $x=a,$ then $f$ is continuous at $x=a .$ Justify your answer.

So let's say f of X represents the cost of a taxi where, um it costs a flat rate of $2 when you enter the cab plus 30 cents for every, um, 1/5 of a mile. So point $30 over 1/5 miles. Um, where x is in miles. So point 3/1 5th That, um if we just multiply that out, that becomes f of X is equal to two. Plus, um, 1.5 x So we know here, this is just a linear equation where $2 is the y intercept, because when you travel zero miles, the flat rate is just $2. And then for every one mile Ugo, it costs an additional dollar and 1/2. So we want no. Is this continuous on the interval 0 to 3 miles. And the answer is yes, it is. Because if we look at this, this is just a linear equation, and linear equations are polynomial, and we know polynomial are continuous. Um, everywhere. Um, also, if we think it won't think about this in the applied to terms with the taxi cab, um, if you go any quantity of miles between zero and three. There's no distance you can travel such that the cost is undefined. If you go 0.1 miles or three miles or 2.999 miles, there's always gonna be a corresponding cost. There is no value on this, UM graph where X, where an X value results in an undefined Why value so it is continuous on 0 to 3 inclusive.

Really given problem, we want to consider the function F of X equals execute minus six X minutes 12. And it's going to be over the interval from 3-4. So if we look at f of three and f of four we see we go from a negative to a positive value. So clearly this is gonna be close, this is going to be closer to the negative or to the three value. So we're going to do 3.3 And then I'm actually make the positive. So 3.1 is going to be more likely. We have 3.2 it's positive. U- 3.15 that goes positive. So 3.13 is better. 3.14 Is positive as well, so we're going to do 3.135. Okay. 3.134 is most likely going to be our estimate. Um in this case we'll get 3.135 it'll round closer to that value. So that's our final answer.

Okay, so we're going to use the intermediate value theorem to show That there is a zero in this function on the interval from 1-2. So the intermediate value theorem says that if we have a value on the interval, say the intervals from A to B. If we have a value of F F. A, that is greater than C, where C is just a constant, so it's equal to some value. And then we have at F f B, we have that F B is less than C. Then as long as this is a continuous, smooth curved function, which this is, then we're guaranteed to have a value at some um value in between A and B. I'll just call it F F. D where D. Is, and I'll even write it down where A is less than D, which is less than B. We're guaranteed to have a value here, that's equal to see. So really, what we need to show is that there is um that F of one is greater than zero or less than zero, since this works either way, and then F of two is the opposite. So if F of one is greater than zero and ff two needs to be less than zero, and ff of one is less than zero than ff two needs to be greater than zero. Then we can guarantee that there is a zero in between that interval. So let's look at F of one, This is equal to four times one of the third minus two times one squared minus seven, which is equal to four minus two minus seven, which is equal to um 2 -7, which is equal to negative five. So now we have that f of one is less than zero, so now we need to to be greater than zero to guarantee that we have a zero in that interval. So it's like that F F two, this is equal to four times two to the third minus two Times two Squared -7. So to to third is eight and if we multiply eight x 4 we get 32 and then two squared is four minus two is eight, and then we have minus 7, 30 to minus eight is 24. And then if we minus seven we get Um 17, so this is equal to 17, so therefore FF two is greater than zero. So if we use the intermediate value theorem, that means that on the interval, so on the interval Um from 1 to 2 we are here guaranteed. Yeah, to have FFC equal zero at some see value. Um in this interval, so as long as C is between one and two, since f of one was less than zero and ff two is greater than zero were um guaranteed to have some value or to have f f c where C is some value equal zero in between that interval.

To the given problem, we're going to First verify that this graph has a zero, so it's going to be x cubed -4 Acts Plus two. Without looking at the graph, we're going to look over the interval from 1-2. So over this interval we see that um it goes from negative 1-2. So because of this um that shows us that we are in fact going to be having a zero somewhere in between here. So we can estimate this by looking more closely. If we bring this down at 1.5 75 we see it's gonna be somewhere in between here and we can keep increasing this until we get closer. Um and we see that we're going to end up getting an approximation that gets very close to zero. If this comes down to 1.7 we see that's gonna be somewhere in between here but we want to get to three decimal places. We end up seeing that that zero value that we get Is going to be 1.675. That's our final answer.