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Let $f(x)=\cos x+3 x$(a) Show that $f$ has a differentiable inverse.(b) Find $f(0)$ and $f^{\prime}(0)$(c) Find $f^{-1}(1)$ and $\left(f^{-1}\right)^{\prime}(1)$

$=\left[\frac{1}{3}\right]$

Calculus 1 / AB

Chapter 3

Derivatives

Section 8

Derivatives of Inverse Trigonometric Functions

Differentiation

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section 3.8 Problem number 29. They give us a function f of X coastline expose three x show that this function has a different chewable inverse. Okay, so we were given a theorem here. Ah, in your book, I think it was the're, um three. And if if f is defensible, So we had the original function. If it's defensible on an interval, then and we know that if F prime of X does not equal zero on that interval, then that's a basically a way of telling us that we know if the if the derivative is not zero, it cannot be 1 to 1. Then we know that f inverse exist and is differential. Okay, so all I really need to do here is find the derivative and show that it's never zero. Okay, so f prime of X is going to be equal to minus sign of X plus three. Okay, let's try setting this equal to zero. So if a negative sign of X plus three is equal to zero, that means that the sign of X is equal to three. Well, this cannot happen. We know that the sign is a number between negative one and one. So this has no solution. So because F prime is never zero, then I know that the inverse exist and that in verse is defensible according to the theory. So again, all I did was find the derivative showed the derivative cannot be zero. So if the derivative cannot be zero, then I'm never gonna have a scenario like this because that's where the river we zero. So I'm never gonna have a function that is not one toe one. Um, so I know that it will have an inverse, and this inverse will be differential. Now we're asked if I ng f zero and f prime of zero So again, what was our definition? F of axe? So at f of X is equal to co sign X plus, was it three x co sign X plus three X and then we had that f prime of axe is equal to minus sign of X plus three. So f of zero is going to be one plus zero, which is one and then f prime of zero is going to be zero plus three, which is three now in part. See who has a couple things find the inverse so f inverse of one. So what do we know? We know that f of zero is one. So from from part B, we found out that f of zero is equal to one. So then, by definition, by definition, off f inverse, we know that f inverse of one is equal to zero. Now, a little bit more work we're gonna have to do for the next part. We're gonna have to find f in verse, so f in verse, prime of one. So to find FM verse prime of one. So let's start with what we know. We know that why is equal to co sign of X plus three x So that's what we know about the function now to find the inverse. What do you do with in verses? You switch X and Y So let's switch X and why? To get the behavior of the inverse. So the inverse is going to be ex equal co sign of why plus three why? And so now let's differentiate this with respect to X. So when I differentiate this so what I'm gonna do with this is use implicit differentiation to find the derivative with respect to X and I get one is equal to So that's gonna be minus sign of Why do I d x plus three? Do you? Why? Dx So this tells me that D y d x times three minus sign A Why is equal to one? So do I. D x is equal to one over three minus. Sign off. Why? And so then what am I trying to figure out here? Um, trying to figure out then if I'm looking at, um So we're down to almost there. We've got d Y d X is equal to one over three minus sign. Why one over three? Minus a sign of why now what we know already. Um, we know that, um, if we look at the derivative, um, we know that f prime. So in this case, we have d i. D. X is one over three minus sign. Why? So we know that when y is equal to zero, this just simply evaluates to 1/3. Okay, so we find that f in verse prime of one is equal to 1/3

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