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Sketch the graph of $f(x)=x|x|$ . Then show that $f^{\prime}(0)$ exists.

See graph slope of the tangent line at $x=0$ is zero, which means that $f^{\prime}(0)$ exists and $f^{\prime}(0)=0$

Calculus 1 / AB

Chapter 3

DIFFERENTIATION

Section 2

The Derivative as a Function

Derivatives

Differentiation

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Okay, so this right here is the graph of F of axe. Cols X times absolute value backs And it wants us to show that the derivative not only exists at X equals zero, but also wants us to find that value. So how do you show that something is differential? Well, just like limits and continuity, the right derivative has to equal the left derivative. So we're gonna end up making this function in the two pieces a right piece in a left piece and show that they have equal derivatives. So how do we do that? Well, we'd really like to get rid of that absolute value. So we'll just consider each case whether X is positive or negative and see what our function turns into. So So let's say f of X at positive values is well, the ex stays on change. The axe is a positive number, and the absolute value is also a positive number. So positive x times positive X is X squared x greater than or equal to zero. But then, if we consider the negative case, we get a negative x times the absolute Valley of Negative X, which is acts so negative x Times X, which is negative X squared if X is less than zero. But we could also put a equal to sign here because F. Of zero is still zero in both of these. So now let's consider the derivative piece wise. So F Prime of X is just the derivative of each piece. So the derivative of two X squared is positive to X. If X is bigger than zero or the derivative of negative X squared is negative to X effects is less than or equal to zero. So to show that it's differential everywhere, all we need to do is show this boundary 0.0. So f prime of zero from the positive direction is two ex evaluated at zero, which is just zero and f prime of zero from the negative direction. This two x negative two ex Sorry, evaluated at zero, which is again zero. So since f prime of zero plus equals f prime of zero negative zero f prime zero exists and it has a value of zero. And again, just to confirm, we can look at our original graph and see that the tangent line at zero is just horizontal. So our intuition matches

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