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1. Use the graph to fill in the blanks be as accurate as possiblef(-1)=f'(-1) ='(-1)...

Question

1. Use the graph to fill in the blanks be as accurate as possiblef(-1)=f'(-1) ='(-1)

1. Use the graph to fill in the blanks be as accurate as possible f(-1)= f'(-1) = '(-1)



Answers

Use the graph of the following function $f(x)$ to find each value.
$f(1)$

We are given the calf off the function on we have to find f off minus one, which means when X equals minus one function as equal do zero Hanta answer is zero.

When trying to determine whether we have a removable discontinuity or a vertical ascent, um at X equals negative one. For this function we can take the limit as X approaches negative one and depending upon the result, we'll know the answer. For example, if we end up with a value divided by zero, that indicates that we have some kind of infinity around negative one. Given us a world class into. Or if we end up with 0/0, which is the case in this particular problem, we plug in negative one. We can see that in fact we have 0/0. This is an indication that we have a removable dis continuity. We further verify this by factoring the numerator and denominator if necessary, dividing out anything that is a common factor and finding our new limit based on this simplified version of the function. So in this case the limit as X approaches negative one gives us the value of negative to this indicates that we have a removable discontinuity at X equals negative one rather than a glass, too.

In this problem, we're gonna look at the function F of x equals the square root of one minus X and then minus one at the end. So we're gonna start with our parent function the square root of X. Whenever I have the square root of X. In my function, it's always going to start with this shape as we make changes to our function. It also changes our graph. The first change I'm gonna make is I'm going to make the the X. Negative and it's inside or under the radical. So this is very important where that negative symbols app when the negative is right next to the X and under that radical it is going to reflect my graph over the Y axis. The next thing I'm going to change is I'm gonna put a one in front of the negative. So this is really 1 -1. And let's see how it changes when that is added. It moves my Graph one unit to the right. And then the final change is the -1 at the end and the -1 is not inside the radical, it's outside of it again. This is very important whenever i subtract one at the end without it being in the radical, It moves my Graph one unit down, so I'm going to get rid of these other three. And this is the correct graph for our function And notice that it looks like it starts at -1 -1. It does cross the x axis at 00 and going left to right left to right. It is decreasing until it gets to that what looks to be maybe an ending point.

In this problem, we're going to take a look at the function F of x equals negative square root of one minus X And then -1 at the end. We're gonna start with our parent function, the square root of X. Any time that you graph the square root of X. It's going to have this shape and then we make changes to our equation and that will change the shape of our graph. The first change we're going to make is we're gonna make it a negative X instead of a positive X. It's very important that this X is inside the radical. So the square root of negative X. Is going to cause our graph to reflect over the Y axis. The next change is there's a negative in front of our radical. So we're going to assume that it's going to do something different because it's placed in front of the radical instead of inside. So once we do this, it reflects our graph over the X axis. Now we have added a one underneath the radical. So instead of it being the square root of negative X, it's now the square root of one minus X. And the fact that the ones inside the radical is very important. It's going to cause our graft to shift one unit to the right. The final change on this function is the -1 at the end and the -1 is outside of the radical. So it's going to do a separate shift. It is going to shift the Graph One unit down so I'm going to get rid of these other four. This is the correct graph for our function notice going left to right. It is increasing. We also have a Y intercept located at 0 -2 and it looks like it stops at 1 -1.


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