5, Usin9 the de finition 2 the lervatiie 3 function 4 Point, Calculak S (-2) eay Sc) = Xt3 -X6. Using the Refi-tion 0 te deri Vative # functibn a} Poit) Calculale 9...

$5-6$ mplete the table of values, rounded to two decimal places, and sketch a graph of the function.
$$
\begin{array}{|c|c|}\hline x & {f(x)=2 e^{-0.5 x}} \\ \hline-3 & {} \\ {-2} & {} \\ {0} \\ {0} \\ {1} \\ {2} \\ {3} \\ \hline\end{array}
$$

So in This problem we're looking for is we're creating a table of values to see what a particular function using the natural base looks like. So in this case, the equation that we're gonna the table that we're gonna work with looks like this. We're evaluating the function three e to the X power. So we're gonna plug in some values. Negative to negative one negative half. Zero have one and shoe. Okay, when I substitute negative to into this function, I'm gonna get zero point for one rounded to two decimal places when I plug negative one. And I'm gonna get 1.1 zero when I plug in negative half. Gonna get one point age, too. When I plug in zero, I'm going to get three. When I plug and a half, I get 4.95 When I plug in one, I get 8.15 and when I plug in to I get 22.17 now, each of these air points Now that I can graph at negative two, it's gonna be a little bit less than a half at negative one. It's gonna be a little bit more than one at 0.5 is gonna be almost at two. At zero. It's going to be at three and a half. It's going to be a little bit less than five at one. I'm going to have 8.15 so it's a little bit more than AIDS. And finally, a two. It could be 22 which is way, way, way up here now on. We know based on our exponential work, that this is going to level off at zero and then get really, really, really steep as it approaches infinity. So here's what we have. The sketch of the graph here, the points.

Is from the number 19. We have three functions I for fax, G, F. X and H. F. X. And we want to find affection value at half of three. Okay. At X equals 23. So now let's plug straight into the three functions. Okay, So F of ah Half of three equals 2. Three to the. Okay. Third power. Okay. It goes to three times three times three, Which is 27 times three, it goes to 81. Okay, So we get half of three equals to 81.

Evaluating this function of three variables here. We're going to replace X with three in the first example. Why with eight and then Z with two. And since it's easy to the negative to, that just means it's squared and then in the denominator. So let's take care of that right away to square it is four, and then negative means it's in the denominator. So now this is the same thing. 24 divided by four would give us six so that first one just evaluates to six. The next one here is three times negative, too times again. Negative six. But to the negative, too. Let's do that so that to the negative to first off we square negative six, which is positive. 36 negative means we put it in the bottom so that would be over positive. 36 because it was negative square. And then we have negative six up top divided by 36 in the denominator, and so that ends up giving us negative 16

Today we're going to solve a problem about you know f of X equal to X cube, minus X squared minus 66 plus two. So it is a poor, normal function. It is always different, troubled and continuous. A false zero equal Do Who f off three Equal to three cube minus three square, minus six in the tree, plus two with physical toe So a 40 ricardo F off three. Hence throws terms applicable in closing double zero comma three differentiator. For fix F dash X equal to three X squared minus two x minus six F D I. C equal to zero three c squared minus two C minus six equal to zero they get. If it's all this court activation, we could see equal to one plus or minus. Route 19 by three one minus Root 19 by three Does not belongs to close intervals. Zero Command three. So hence it is rejected. So she is a called a one plus route 19 by three. Thank you