Question
Use the graph of f (c)below to find a number 0 such that Iz2 _ 1| 0.1 whenever 0 < Ix - 1| < 6.1.10.9Round your answer to four decimal placesNumber
Use the graph of f (c) below to find a number 0 such that Iz2 _ 1| 0.1 whenever 0 < Ix - 1| < 6. 1.1 0.9 Round your answer to four decimal places Number
Answers
Use the graph of $f$ in Exercise 9 to estimate all values of $c$ that satisfy the conclusion of the Mean-Value Theorem on the interval [0.4].
Were given a function f and rescues the mean value theorem and find all the points. See between zero and 2. Such the effort to minus F of zero equals at private see temps to minus zero function is that F of x equals X -1 To the 10th power. No, I don't miss my boyfriend now this is the derivative F prime of X while F is differentiable and the derivative is 10 times x minus one to the ninth power. Using chain rule. They are let's watch a bunch of. Therefore, by the mean value theorem, it follows that there exists at least one C in the open interval 02 Such that f of two -F of zero equals F. Prime of C times two minus zero. To plug in to to our expression we have F of two is one of the 10th or one minus F of zero, which is negative one to the 10th, which is just one Since 10 is even. This is equal to f prime. FC is 10 times C -1 to the ninth power times. To therefore you find that c minus one to the ninth equals zero, or in other words, C equals one.
No given a function F and the rest is the mean value theorem and find all the points see between zero and two. Such that F of two minus F of zero equals f. Prime of C, attempts to minus zero. The function F of X Is X -1 to the 9th. Mm hmm. Actually, now this function is differentiable and the derivative at prime of x. But the chain rule is nine times X -1 to the 8th. Therefore, by the mean value theorem, we know that there exists at least one C in the open interval zero to such that F of two -F of zero equals f prime of C times two minus zero plugging in. This is one to the ninth or one minus negative one to the ninth, which is negative one since nine is odd, this is equal to nine times C -1 to the 8th times two. Therefore we have a C -1 to the 8th is equal to 19 Therefore it was fucking c minus one to the fourth is equal to plus or minus one. Third Of course this is positive. So it follows AC -1. The 4th can only school to positive 1/3. Your honor. I just want to thank. But then it follows That C -1 is equal to well C minus one squared is equal to plus or minus one over the square root of three. But again this is positive. So that c minus one squared is only equal to the positive one over square root of three. And therefore We get C -1 equals plus or -1 over the square root of the square roots Or the 4th root of three. And therefore C is equal to one plus or minus one over the fourth root of three. Since the magnitude of one over the fourth root of three Is less than one follows that both of these seas lie in the interval from 0-2, and therefore see is one plus or minus one over the fourth root of three.
So in this problem were given this graph we're asked to find the number delta. Such that If zero is less than the absolute value of X -3, less than delta then F of x minus two sorry minus two is less than 0.5. Okay, so You can see from the graph right that when F of X -2 is less than .5? Well then we are in this range aren't we? Somewhere between 2.5 and 1.5 for the F of X. Okay. And that means then that We have these two deltas, right? Delta one Will be 3 -2.6 which is 0.4 and Delta two which will be 3.8 -3 which is me fix that up just a little bit right there because that's an absolute value sign there we go. 0.8. And so then that means that the delta we're looking for is the minimum Of these two Delta 1 and delta two. The minimum of those two is .4. So this is 0.4. So we noticed that If .4 is delta and everything less than .4 away from three, Right from 2.6 up to 3.4 being that range right there wouldn't It is all within 0.5 of two on the f. Of X, is it not? It is okay. Yeah so there's our answer. Um Right there
This question asked us to use the graft to estimate the values of see that satisfy the conclusion of the mean value thehe. What we know, given the graph, is that we can find approximate values of the seats. In other words, we can see access 4.5 and X is 6.1. We can see a tangent line on the curve. Therefore, these could be considered to be our C one and R C two.