💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here!

Like

Report

Numerade Educator

Like

Report

Problem 79 Hard Difficulty

Prove Fermat's Theorem for the case in which $f$ has a local minimum at $c$.

Answer

If $f$ has a local minimum at $c,$ then $g(x)=-f(x)$ has a local maximum at $c,$ so $g^{\prime}(c)=0$ by the case of Fermat's Theorem
proved in the text. Thus, $f^{\prime}(c)=-g^{\prime}(c)=0$

Discussion

You must be signed in to discuss.

Video Transcript

we are going to prove Fermat's theory for the case in which the function F has a local minimum at sea. So we know that in the text book, it was proved the theorem for the case in which the functions has local maximum. And we got to remember that the theorem says that if the function has a derivative at the point Where it has local maximum and the serial zero. So um we want to prove that the S. F. Has local minimum. And see then conservative At that point if it exists is zero. But for these we're gonna use exercise 78. Which says that if a function F has a local minimum at the point C then G F X defined as negative Fx has local maximum value at team at the same point. So here we're in that case because we are supposing that F has a local minimum at sea. That's the hypothesis for the exercise 78. So, directly from exercise 78 we know that the function G. Fx define its negative F. Of X has local maximum at sea. Mhm. Okay. So now because she dysfunction has a local maximum at sea, we can apply the part of the firm and theory. That was proved in the textbook. That is the case of the local maximum at a point. And we know the function G has local maximum at a point C. So you send him and part of the theory. Okay, proof into text. Mhm. We know that G. Tell you that in that C. Is zero if you exist. So if the function is there has a relative function G. Has already at sea, we know that F also has the relative at sea because we know that she's negative F. So the only difference is a sign. So differential ability. One of the functions implies the differential ability of the other. And more over than that. We know that the derivative of F is equal to negative the derivative of the X. For any X. Because this function here again, but the negative sign to the left of the equation, an F. Is equal to negative G. And derivative of F is equal to negative. But this equation holds at sea. There is a derivative of F. At sea is negative period of GFC, but it's zero because of this. And this implies that the derivative of FC0. So, yeah. First is here theory has been proved, yeah. For the case of a function F. That has local millions. The idea was using what was proved in the textbook. That is the case for the local maximum at sea. And the exercise 78 that says that if the function has local minimum at a point in negative, that function, this new function to finance negative the given function as a local maximum. Using then the theorem, the part of the theory that was approved in the test book, we concluded the derivative of the given function at sea 8 0. And then this is the proof of the theory in the case of the local minimum adequacy.