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Find the largest possible subset A of R that will make the following functions well- defined.2f:A-R given by f(x) = V3x+15x+3b. f:A-R given by f(x) V(x+2)(4-x)...

Question

Find the largest possible subset A of R that will make the following functions well- defined.2f:A-R given by f(x) = V3x+15x+3b. f:A-R given by f(x) V(x+2)(4-x)

Find the largest possible subset A of R that will make the following functions well- defined. 2 f:A-R given by f(x) = V3x+15 x+3 b. f:A-R given by f(x) V(x+2)(4-x)



Answers

a. Find the open intervals on which the function is increasing and those on which it is decreasing. b. Identify the function's local extreme values, if any, saying where they occur. $$f(r)=3 r^{3}+16 r$$

Okay, so four problem 25 way have here is our which is divine to be three are huge loss 16 are get now for part A We need to take that the rift here off this function So that gives ny r squared a 16 response to we let this to be positive Now the solution will be ny r squared, eager than active 16. But this is always true and we let hard to be. Well, this is always true because R squared is always bigger. We put 20 so there are times zero times nine will be zero and zero is still bigger than 16. So that means that far is increasing on are on the real numbers now for part B. So since we conclude that the function is increasing, its always increasing on on the real numbers, so there's no extreme values

So we have the function h of r equals R plus seven to third. And we want to find where this function is increasing and decreasing and where any local extreme a occur. So do that. We'll find the first derivative h prime of our using the chain rule we get. The derivative is three times are plus seven squared, sometimes a derivative of R plus seven, which is just one. We'll leave that out, and now we want to find the critical value of each of our and the critical values of a function are where the first derivative is equal to zero or where it's undefined because this is a quadratic, um, three times are plus seven squared. It is continuous everywhere. It's never going to be undefined. So just look for when it's equal to zero. So if we do viable sides by three, we get our plus seven squared equals zero. Take the square root herbal sides we are plus seven equals, um, zero again. So we get our equals negative seven. So when r equals negative seven h prime of our is zero. And so if we set up a number line where? Right here we have negative seven. We want to find how h prime of our behaves to the left and to the right of this critical value at r equals Negative seven. So we'll just use a test value. We'll use the test value, say, with negative 10 to the left of it. Um, so if we plug in negative 10 to h Prime of are we get negative 10 plus seven, which is negative. Three squared is nine. Um, Times three is 27. So that's a positive number. So we know that to the left of our equals negative seven, the numbers are all going to be positive. And we know there are always gonna be positive because if they were ever to switch back to negative than over here could be another critical value. Where, um, it equal zeros that could go back. It could start being negative again, but because it never, um, there's no other critical values. We know that it's Onley positive values to the left. Similarly, we can use a test value to the rights we can use the value of zero. We plug in zero th prime of are we get zero plus seven square that's just seven squared 49 times three. Whatever that turns out to be, that's also going to be a positive number. So we know that both to the left and the right of the critical value, Um, the first derivative is positives. Was that mean well, when the derivative, the first derivatives of the function is positive, means that the function is increasing on that interval? So because it's increasing both the left and the right of that critical value, we know it is increasing everywhere, So it's increasing from negative infinity to positive infinity because it's increasing everywhere and it never turns a corner. If we can picture the function, it's gonna be increasing everywhere and then at that it increases again, increases again, so there's no extreme values. There's no minimum or maximum because it just keeps going up higher and higher, lower and lower in the other direction. So we know there are no local extreme a and this makes sense because if we think about just the original function, age of R equals R plus seven to the third Power. This is just a cubic function with a horizontal shift to the left, seven units. This is what that function is. Our plus seven to the third power that looks something like that, where at r equals negative seven. At this point, the derivative is zero because the slope at that exact point is zero. But even though it zero, there still isn't a it doesn't change directions. It doesn't start decreasing from there. So there's still no local extreme a and it is everywhere increasing.

Okay, so here we have. Each of our is our plus seven Cubans. We take the derivative we get, three kinds are seven squared. And if we want to find our critical points for each count zero, well, that's just gonna be in our equals. Negative and every troll number line. See where functions increasing or decreasing. We have made seven will test. Yeah, it's prime. But H crime is always greater than equal to zero. So each crime is going to be positive over here and positive over here. So age is going to be increasing from minus infinity to mine seven. And it's also going to be increasing from seven to infinity, and so they're actually no local. For absolute extreme. This function just increases and the levels often and increases like that.

In this problem of relation and function we have to prove that the greatest interior funds and F, which is different from our tour and it's given as FX is equals the greatest integral part of X is neither when we're nor onto. So for this first we will check for 11 So here check for 11 So a function is said to be 11 If F of excellence is equals two. F of X two implies that Excellent. And this will be excellent will be equal to X two and provided that excellent. And next to would be from the given domain. So that means here Excellent and X two belong from the set up. And now we are taking say F of X one. Say this would be here F of Excellency. Excellent is equal to 1.2 and say X two is equal to 1.5. So when we write it F of excellence. So this would be F of X one. And here it will be F of X two. From here. F of X one. That means greatest integral part of excellence. So this will be the greatest interior part of 1.2 and this should be integrated into your part of 1.5. So from here we conclude that the city equals to one. And from here we conclude that the city equals to one. So when we see F of X ray energy equals two F of X two. But here excellent is not equals two X two because 1.2 cannot be equals to 1.5. So that's why we conclude that the function great in the interior part of X is not 11 So we say that difference and F is not 11 And now we have to check for on defense. So here this will be from our to our So this will be the domain. See this will contain values like 1231.51 point six 1.98 C. Class. But here this would be ethics C. X. And here this would be value of life. See why is equals two fx. And here and some values will be like 1231.51 point 62.1. And the values like 1.5 1.6 and 2.1 would we left behind because why is always in teacher because this is defined as why is equal to british interior part of X. And this will give you only interior part. So that means this value doesn't have any pre image in X. So that's why the funds and F or we can say greatest interior part of X is not on to function. So this is the answer that function is or we can say the greater into your fans and neither 11 or or two


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