Question

At what numbers is the following function $g$ differentiable? $$ g(x)= \begin{cases}2 x & \text { if } x \leqslant 0 \\ 2 x-x^2 & \text { if } 0<x<2 \\ 2-x & \text { if } x \geqslant 2\end{cases} $$ Give a formula for $g^{\prime}$ and sketch the graphs of $g$ and $g^{\prime}$.

   At what numbers is the following function $g$ differentiable?

$$
g(x)= \begin{cases}2 x & \text { if } x \leqslant 0 \\ 2 x-x^2 & \text { if } 0<x<2 \\ 2-x & \text { if } x \geqslant 2\end{cases}
$$


Give a formula for $g^{\prime}$ and sketch the graphs of $g$ and $g^{\prime}$.
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Single Variable Calculus: Early Transcendentals
Single Variable Calculus: Early Transcendentals
James Stewart,… 9th Edition
Chapter 3, Problem 76 ↓
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At what numbers is the following function $g$ differentiable? $$ g(x)= \begin{cases}2 x & \text { if } x \leqslant 0 \\ 2 x-x^2 & \text { if } 0<x<2 \\ 2-x & \text { if } x \geqslant 2\end{cases} $$ Give a formula for $g^{\prime}$ and sketch the graphs of $g$ and $g^{\prime}$.
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Transcript

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00:01 Okay, so we are asked where this function g, which is the function that i have graphed, is differentiable.
00:11 So let's first look on the interval negative infinity to zero.
00:19 Now, we just need to check to see if the function is differentiable on zero.
00:25 So we need to make sure that the derivative coming from the left at zero is going to be equal to the derivative coming from the right.
00:32 At zero.
00:33 So the derivative coming from the left, it will call f prime plus is going to be equal to two.
00:40 The derivative coming from the right is that we'll call f prime minus x is going to be two minus two x.
00:55 So now we just want to evaluate both of these at zero and see that they're equal.
01:00 So this is going to give us 2 equal to 2.
01:03 So the function is differentiable at 0.
01:06 So that's our first interval.
01:08 Now we want to look from 0 to 2.
01:16 And let's see if it's differentiable at 2.
01:19 So let's look at the derivative from the left, which is going to be 2 minus 2x.
01:26 And then the derivative from the right, which is going to be minus 1...
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