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[et f be a real- -valued function defined on (_o, such that | f (x) Isler - 1| for all x in (-o, 0 ). Compute the value of f (0) and determine whether f is continuo...

Question

[et f be a real- -valued function defined on (_o, such that | f (x) Isler - 1| for all x in (-o, 0 ). Compute the value of f (0) and determine whether f is continuous at 0. 8 marks)Let fbe a real-valued function that is continuous o [0, 1] and differentiable on (0, 1)and let g be a function defined on [0, 1] by g (x) =f (x) - (Y) x Iff(0) = 0,f ()-and f (1) = 0,show that there exists € in (Y, 1) such that g (c) = 0.(6 marks)(ii) determine whether the equation g (x') 0 has any real root in

[et f be a real- -valued function defined on (_o, such that | f (x) Isler - 1| for all x in (-o, 0 ). Compute the value of f (0) and determine whether f is continuous at 0. 8 marks) Let fbe a real-valued function that is continuous o [0, 1] and differentiable on (0, 1) and let g be a function defined on [0, 1] by g (x) =f (x) - (Y) x Iff(0) = 0,f ()- and f (1) = 0, show that there exists € in (Y, 1) such that g (c) = 0. (6 marks) (ii) determine whether the equation g (x') 0 has any real root in (0,1). Justify your answer: (6 marks)



Answers

Assume that $f$ is differentiable for all $x,$ where $f^{\prime}(x)>0$ on $(-\infty,-4), f^{\prime}(x)<0$ on $(-4,6),$ and $f^{\prime}(x)>0$ on $(6, \infty)$ Supply the appropriate inequality symbol for the given value of $c.$ (Sign of $g^{\prime}(c)$) $$\begin{aligned}&g^{\prime}(0)\quad 0\\&g^{\prime}(-5) \quad 0\\&g^{\prime}(-6)\quad 0\\&\begin{array}{lr}g^{\prime}(0) & 0 \\g^{\prime}(0) & 0 \\g^{\prime}(8) & 0\end{array} \end{aligned}$$ (Function) $$g(x)=-f(x)$$

So for the following problems were considering a case in which we know that G has a relationship to F. So G F X is somehow related to F of X. Whether that's through some multiple, like negative one or three or something like that. And then in some cases we may add a constant as well. But notice that when you take the derivative G prime of X. The only thing that matters is not the constant is just going to be F of X. But you're gonna have F prime of X and then times whatever multiple. We had some constant. A. So when we consider that we want to keep in mind that F prime of X is going to be greater than zero on negative infinity to negative four, less than zero on negative 4 to 6 and greater than zero from six to infinity. So based on whatever this is and whatever our multiple is, we'll be able to know if G prime of the values zero or eight or whatever Is greater than zero. Yeah, or less than zero. So that can be how we find the final answers

So for the following problems were considering a case in which we know that G has a relationship to F. So G F X is somehow related to F of X. Whether that's through some multiple, like negative one or three or something like that. And then in some cases we may add a constant as well. But notice that when you take the derivative G prime of X. The only thing that matters is not the constant is just going to be F of X. But you're gonna have F prime of X and then times whatever multiple. We had some constant. A. So when we consider that we want to keep in mind that F prime of X is going to be greater than zero on negative infinity to negative four, less than zero on negative 4 to 6 and greater than zero from six to infinity. So based on whatever this is and whatever our multiple is, we'll be able to know if G prime of the values zero or eight or whatever Is greater than zero. Yeah, or less than zero. So that can be how we find the final answers

Hello were to determine function is continuous urgency fr works. That is the costume X squared minus six X. No, Upon excess surplus six x. If X is not a question zero And it is -1 or exit goes to zero. So the value of the limit extends to zero of fx. So this is the cost to 96 a pound six. So this will be a costume minus one. Okay? And the f of zero is -1. So we can see that limits Extents to zero of FX. Here's your costume on this one, your costume F of zero. So we can see that the function is continuous. Function is is continuous. Okay. I hope you understood.

For this problem first, let's verify the intermediate value theorem applies here. So, for the intermediate value theorem to work and must meet the following three criteria. First, the curve is the function Y equals F of X. Now, if we look here, we do have an fx that's given by x squared minus six X plus eight second, which is continuous on the interval A. B. The function must be continuous. Well, our function is a second degree polynomial which is continuous for all real numbers in its domain, so 0 to 3. That'll work just fine. And W is a number between F of A and F A B. This is what we must verify for this problem. Now, W is going to be given by this number right here. FFC is equal to W. So are W number is zero, and it must be between F zero and F three. If it doesn't fall between these two numbers, then the intermediate value will not. The intermediate value theorem will not apply here. So first it solved for F zero, F zero is going to equal zero squared minus six times zero Plus eight, which is just eight. Now, F F three is given by three squared -6 times three plus eight. Which is just don't worry about that. Nine minus 18 plus eight. Or It's gonna be a -1. And so let's check that zero fall between negative one and eight. Yes, it does. Doesn't matter if F zero is the bigger number of the smaller number, as long as the value W which is zero, of course falls between the two. So it will work. And we can prove that FFC does coach W. So there is some value. See that when plugged into this function will give us an output of zero. So let's find it. Well, let's first try Our other integers so we can try f of 1 1st. And that's going to give us Spirit. Mind, did the function up here, it's going to give us one squared -6 times one plus eight. And when we saw for that we'll get one minus six negative five plus eight. That's going to give us a positive three. So that's not quite it. Let's try to Now, since that's the only other integer in our given domain, two squared minus six times two plus eight. Which is going to give us four minus 12 Plus eight. And that's simply going to be four four plus eight minus 12. That's gonna be zero. So we can prove that C is equal to two, which we solved four by plugging in. And we proved with the intermediate value theorem that it is possible that it's guaranteed that there is a value of C. Such that zero would have been an output.


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