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D (3f(x)) dxSelect one:a3 f(x)b 3 F(x) f2(x)C2 f (x)di 1 3 F (x)...

Question

D (3f(x)) dxSelect one:a3 f(x)b 3 F(x) f2(x)C2 f (x)di 1 3 F (x)

d (3f(x)) dx Select one: a 3 f(x) b 3 F(x) f2(x) C 2 f (x) di 1 3 F (x)



Answers

Evaluate the function for the given values of $x .$ $$f(x)=\left\{\begin{array}{ll}x^{2}-3 & \text { for } x \leq 1 \\2 x+1 & \text { for } x>1\end{array}\right.$$ a. $f(3)$ b. $f(0)$ c. $f(1)$ d. $f\left(\frac{3}{2}\right)$

So here we have been given that integral of F X. Dx is equal to have effects and we need to figure out the value of the integral F F X squared times Dx. So for this we have to remove the integral of this given statement. So in removing this we get F of X. This term is equal to du by dx off F of X. So from here we can again rearrange and we can write it as D off affects divided by ffx. That will be equal to dx. So now we integrated. So while we integrate on the right hand side we're going to get X plus C because integral of dx X. So we'll get X plus C where C is some positive constant or any constant an integral of D of effects by effects. If we have to integrate this D of fx by fx we're going to make this substitution here. So let's take care of of access T. So this integral will change to integral of DT by T. So that will be lofty. So alan represents like to base E so that will be basically log off fx to base E. So here after integrating weekend, this has log off. Fx. Alan represents like too busy. So here we're going to write this as we can make a small change this constant. Of course we can write it as LNC as well. So we get alan of F X is equal to X plus LNC and X. We can rewrite it as Ellen E to power X. So we get the expression as Ellen F of X is equal to Ellen E two part X plus LNC. So we know that log off sometimes and is log simples lock and so we use this identity and against simplified the statement here. So we get log off. Fx is equal to log off. C times E to bar X. So we remove log from both sides and we can F of X as see time's Tito barracks. And from here integral of FX whole square times T X will be equal to integral of FX whole square will be C square feet apart two X times DX. So from here, see square will come out of this integral an integral of each part two X dx. That would be A two Part 2 x over two. So basically we see that F of X whole square will be c square feet apart two X. So this is equal to affects Whole Square by two. So this is the result of the required statement

Here. We're in a plug in negative one and native one plus three is too. Plug in to third, so that gives us three in to third. So that's going to give us three plug in 1.3. That's gonna get me 4.3 and my answers before.

For this problem. We've been given a function f of X equals X plus one divided by two X minus three. And our goal for this exercise is to, uh, find the value of this function at some given input points. So let's begin. Let's start with F of zero. What does it mean when we cf zero? What it means is we're going to go find our F function, which is our function we've been given and everywhere where an X appears in that function. I'm going to substitute in the value of my parentheses in this case, zero. So if I go to the F function, I'm gonna put in a zero every time I see an X in my function. So if I evaluate this and simplify it, I end up with negative one third. Okay, let's try a different number. Let's do f of negative one again. Go to the F function and substitute negative one end. Now, when I see X, that'll be negative. One plus one over two times. Negative one minus three. Well, the top of that is going to be zero. The bottom is gonna be negative. Five, which means that simplifies to zero. Okay, One more number. Let's put in f of three function F Substitute three and for X. That gives me three plus one over two times three minus three. So my numerator is for and my denominator is three. So f of three is four thirds. Now we've been substituting numbers in for X. You can also substitute expressions things that maybe have variables in it. For example, f of two X. Well, just because there's a variable in those parentheses, the process doesn't change. I'm still going to the F function and where I see X in the original, I'm going to substitute two X so that's going to give me two X plus one over two times two X minus three. And let's get rid of those parentheses That gives me two X plus one over four X minus three. Now, this isn't as nice of an answer. Doesn't look as nice as the numbers. I still have a letter in it, but that's okay, because I input it a variable into my function. Typically, if I input in a variable, I'm gonna have a variable in my output. So they put two X that fractions what we get out. Let's try one more of thes. Let's do F of X plus H well original function Everywhere. There's an X now gets an X plus h, so I have X plus H plus one over two times X plus H minus three. And again, let's get rid of our parentheses. That gives me a denominator of two x plus two H minus three. So for our given function F, here are the results when we evaluate our function at various different values for X.

Again this question, what is given that is integration of effects DX from 1-5. It is given as five And integration affects dx from 1 to 9. That is given three And integration of F x D x from 3 to 9 it is given us four and now we have to evaluate part that is integration of 5 to 9. F x dx. Okay, so 5 to 9 so we can rewrite it from 1 to 9 minus 1 to 5. So 1 to 9, F x d x minus 125 F x dx Ok. And now we will put the values so 1 to 9. It is from here three okay -1-5. From here it is five so 3 -2. It will be 3 -5, it will be -2. That is the answer of part and now part B we have to find out integration of f x dx from 3 to 5 and that will be integration 3 to 5 can be got by 3 to 9 minus 529329 F x D x minus 5 to 9. F x D X. And now we will put the values from the given data. 3 to 9. That is four from here. And 5 to 9 329 and 5 to nine which we have got from the part of the solution. A Okay. 5 to 9 that is minus two. So it will be minus two and four minus minus two. It will be six so six will be the answer of our part A Okay. And now part C that is we have to find out integration of 321 F X D X. Okay, so that can be written as first of all minus integration F x D x. From 1 to 3. And now 123 can be written as first of all minus common factor and 123 can be written as 1 to 9 minus 3 to 9. Okay. 129 f x d x minus 329 fx dx. And now we will put the values so 1293 Okay. And 329 is four, so it will be three minus four. Okay, First of all negative sign and three -4. So it will be -2 -1. That will be positive one and positive one will be the answer of our part. C. Okay so I have mentioned all the answers here. Thank you.


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