Question

Find numbers $a$ and $b$ such that the given function $g$ is differentiable at 1 . $$ g(x)= \begin{cases}a x^3-3 x & \text { if } x \leq 1 \\ b x^2+2 & \text { if } x>1\end{cases} $$

    Find numbers $a$ and $b$ such that the given function $g$ is differentiable at 1 .

$$
g(x)= \begin{cases}a x^3-3 x & \text { if } x \leq 1 \\ b x^2+2 & \text { if } x>1\end{cases}
$$
Single Variable Calculus: Early Transcendentals
Single Variable Calculus: Early Transcendentals
James Stewart,… 9th Edition
Chapter 3, Problem 86 ↓
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Find numbers $a$ and $b$ such that the given function $g$ is differentiable at 1 . $$ g(x)= \begin{cases}a x^3-3 x & \text { if } x \leq 1 \\ b x^2+2 & \text { if } x>1\end{cases} $$
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Transcript

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00:01 There's a couple of things to do in order for g of x to be differentiable.
00:06 And we want to make sure that it's differentiable at x.
00:11 X equals 1.
00:15 Make sure i have the right equation down, vx squared plus 2.
00:22 And the top one's less than or equal to.
00:24 The bottom is x is greater than 1.
00:27 But anyway, the first thing is that you want to make sure that the function is continuous.
00:33 Us.
00:36 So if you were to plug in one in for all these xes, well, one cubed is one, and three times one is three, that needs the equal plugging in one for this x, one squared is still one, and we can stop right there, because what you also need is a function where you're differentiable.
00:57 So i'm going to take the derivative first.
00:58 It'll be 3a x squared, just move the exponent in front, subtract one from the exponent, minus three.
01:07 Needs to also equal the derivative down here, which again, 2bx, and then the derivative of 2 is 0...
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