Problem

Suppose that $ f" $ is continuous and $ f'(c) = f…

03:48

Problem 89 Medium Difficulty

Show that the function $ g(x) = x | x | $ has an inflection point at $ (0, 0) $ but $ g"(0) $ does not exist.

Name: show that the function gx x x has an inflection point at 0 0 but g0 does not exist
Uploaded: 2021-05-14
Duration: 6 min 1 s
Channel: Numerade
Description: Show that the function $ g(x) = x | x | $ has an inflection point at $ (0, 0) $ but $ g"(0) $ does not exist.

Answer

Using the fact that $|x|=\sqrt{x^{2}},$ we have that $g(x)=x|x|=x \sqrt{x^{2}} \Rightarrow g^{\prime}(x)=\sqrt{x^{2}}+\sqrt{x^{2}}=2 \sqrt{x^{2}}=2|x| \Rightarrow$
$g^{\prime \prime}(x)=2 x\left(x^{2}\right)^{-1 / 2}=\frac{2 x}{|x|}<0$ for $x<0$ and $g^{\prime \prime}(x)>0$ for $x>0,$ so (0,0) is an inflection point. But $g^{\prime \prime}(0)$ does not
exist.

Catherine R.

Missouri State University

Caleb E.

Baylor University

Kristen K.

University of Michigan - Ann Arbor

Joseph L.

Boston College

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Video Transcript

So we want to show that there's an inflection point at This zero. However, G double prime of zero, we want to also show that this does not exist of this is equal to DNA or does not exist. And so the way that we're going to do that is we're gonna set the square root or, sorry, the absolute value of X equal to the square root of x square. And then we're only going to take the positive values of the square root and we're then gonna find the derivative. So we do this, we have G. Of X is equal to X times the square root of X squared. We're gonna have to use the product rule to find the first derivative. So that would be the derivative of the first times a second. Which would just be the squared of X squared plus the second times the derivative of the plus the first times the derivative of the second. And the derivative of the second would be we'd have to bring down the one half. We bring X squared to the negative one half and we'd multiply by two X. So this is equal to the square root of X squared plus X squared times X squared to the negative one half, which is also just equal to the square root of X squared plus X squared over the square root of X squared. And so now what we can do is we can find our second derivative. So if you do this G double prime derivative of the square of X squared, we bring the one half out. You put this negative one half and then we multiply by two X. And then in the next one we're going to have to use the quotient rule which is the um derivative of the first, which would be or the derivative of the numerator, sorry, Which would be two X times the denominator, which would be square root of X squared minus. And I'm actually going to move this since I'm running out of space minus the first, which is the squared of X squared times the derivative of what is in the denominator. Actually this was sorry, this is minus X squared times the derivative of the square root of X squared, which would be X to the negative one half. And I'm running out of room again. So I'm actually going to just put this part down here. So Next to the negative 1/2 times, I'm sorry, this wouldn't be X to the negative one half. This would be X squared To the negative one half, times two X. And then times one half. And then this is all divided by the squared of x squared squared, which would just be square root of x squared squared. So now if we simplify this and substitute in our absolute value of X for the square root and this should be X squared to the absolute value of X for the square root of X squared. We can actually look at our second derivative. So this is we get rid of these twos, we get X divided by the absolute value of X, since it's divided by the square root of X squared and plus two X times the square root of x squared, which is absolute value of x minus X squared times two X would be extra third and then two times one half would just be one. So this is X squared, Your X to the 3rd times one over the absolute value of X. And then this is all over the absolute value of X squared. And so now that we have our second derivative in this form, we can actually look at our point of um zero. So if we plugged in zero into this equation we have 0/0 plus 0/0. So this is +00 Let's 0/0 which is equal to does not exist. So we do in fact have our second derivative not existing now. We just have to make sure that it is actually an inflection point. So we go from negative values for a second derivative to positive values are positive values to negative values. So if we plug in this point of negative one We would get negative one divided by one plus negative two times one which is just negative two minus -1 to the third would be negative one. And then we have this Negative again. So this is plus one And then divided by one. So we have this being equal to negative 1 -1 negative two. Plus one is negative one. So minus one, sequel to negative two which is less than zero. So we're gonna want our Value that is greater than zero Or value at g double prime of one to be greater than zero. So if we plugged in one into our original equation we'd have 1/1 which is just one plus two times 1 is 2 -1 of the third is one Times 1/1. So this is 2 -1 in the numerator and the denominator we just have one squared this is over one, so this is equal to one plus one which is equal to two which is greater than zero. So we are going from negative values to positive values. So we do have an inflection point at 00 And we do not have our second derivative actually having a value or does not exist at 00. So we did indeed show what we needed to show

Kian M.

Oregon State University

Related Courses

Calculus 1 / AB

Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 4

Applications of Differentiation

Section 3

How Derivatives Affect the Shape of a Graph