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Point) Suppose f(4) = 6, f' (4) = -2,and dH H(x) = (1 + f(x)) . Find dx X=4 H'(4) =...

Question

Point) Suppose f(4) = 6, f' (4) = -2,and dH H(x) = (1 + f(x)) . Find dx X=4 H'(4) =

point) Suppose f(4) = 6, f' (4) = -2,and dH H(x) = (1 + f(x)) . Find dx X=4 H'(4) =



Answers

Suppose $f(x)=4 x-2 .$ Find $f\left(\frac{x+2}{4}\right)$

With this fact is true that f of X plus one is equal to three times F of X minus two. Then we know that this input is always going one more than this input here. So we also know that F zero is four. So if we think of now, after backless woman's effort one and that would be equal to three times f of zero minus two because this is one more zero so f of one would be equal to three times were given the fact that f of zero is for so we could substitute for over half of zero. Then I am three my three times for minus the richest head. Now follow me. This pattern is gonna continue. We have some of the next step f of two would be equal to three times one minus two again because this input is one more unit than that. So after two would be equal to three times Well, we just found off of one of the step above it is 10. So three times 10 minus two is 30 minus two, which is 28. Okay, where you keep this process going F of three would be equal to three times F of to minus two. And so after three equals three times 28 minus two, which is 82. And now we can answer the question with the last 1/2 afford the equal to three times F of three minus two, and we just found f of three to be 82 so three times 80 to minus two is 244.

So picking we have the second derivative of X. And it's equal to four plus six X plus 24 X. Screwed. So when we evaluate this, when you take F prime of X using the anti derivative we get four X plus um free X squared plus. He acts cute Classy. But we know that F of zero is three. So what we're gonna end up getting is Or f of f prime of one is 10. So we're going to end up getting to as R. C value. Then we have F of X. Which is going to give us um to have squared plus X cubed Class two X to the 4th plus two X plus another constant value. But we know that F of zero is three. That means as constant as three. Therefore our final anti derivative is two X squared plus X cubed plus two X. To the fourth plus two X plus three. Final answer.

My we want to evaluate the given functions half of x Y is equal to x minus six Y. For F we quantified f of zero negative four and negative three to this question is testing our ability to work with multi variant function, which is a precursor to continuing on in multi variable calculus. In order to answer this question, we need to understand that this is just like a single variable function where we're just going to plug in the indicated variables. For instance, for F zero negative four, we call getting zero wherever we see X negative for wherever we see why. Thus we can proceed to F zero negative four. So zero negative four is two times zero minus six times negative for the first time goes to zero and negative six times negative four is positive 24 as boxed in. Now that we understand how to do this, we can do identical steps for calculating half of negative three to have negative 32 is two times negative three minus six times two or negative six minus 12, which is not negative 24 but instead negative 18.

So f of X is for X, and we need to evaluate it for X 0123 and four. So let's plug those in. So f of zero for X zero is four times zero. So that zero half of one is four times one, which is four. Half of two is four times to, which is eight F of three is four times three, which is 12. And finally I have 44 is four times for which is 16. So these are your answers for X zero all the way to four.


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