Question

The third-degree Taylor polynomial for $f(x)=x^3\left[2 \log x-\frac{11}{3}\right]$ about $x=a$ is $$ T_3(x)=-\frac{2}{3} \sqrt{e^3}+3 e x-6 \sqrt{e} x^2+x^3 $$ What is $a$ ? 

   The third-degree Taylor polynomial for $f(x)=x^3\left[2 \log x-\frac{11}{3}\right]$ about $x=a$ is

$$
T_3(x)=-\frac{2}{3} \sqrt{e^3}+3 e x-6 \sqrt{e} x^2+x^3
$$


What is $a$ ?
CLP-1 Differential Calculus 1
CLP-1 Differential Calculus 1
Joel Feldman, Andrew… 1st Edition
Chapter 3, Problem 6 ↓
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The third-degree Taylor polynomial for $f(x)=x^3\left[2 \log x-\frac{11}{3}\right]$ about $x=a$ is $$ T_3(x)=-\frac{2}{3} \sqrt{e^3}+3 e x-6 \sqrt{e} x^2+x^3 $$ What is $a$ ?
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Transcript

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00:01 23 for this function, so we're going to have to use the quotient rule on it.
00:11 It's the bottom times the derivative of the top, minus the top, times the derivative of the bottom over the bottom squared.
00:26 It's the bottom, times the derivative of the top, minus the top, times the derivative of the bottom, over the bottom square.
00:35 So now we have minus x, minus 2x plus 2x l in x, or x to the fourth.
00:55 So minus 3x plus 2x lnx over x to the fourth.
01:01 Let's go ahead and take an x out of everything.
01:03 So we have minus 3 plus 2 lnx over x cubed.
01:12 All right, next.
01:15 Bottom times the derivative of the top minus the top times the derivative of the bottom over the bottom squared that gives us 2x squared minus minus so plus 9x squared minus 2 times 3 so minus 6 x squared l and x over x to the 6 so 11 x squared minus 6x squared lnx over x to the 6th.
02:00 So 11 minus 6 lnx over x to the 4.
02:06 Okay, i thought i would be able to guess better what the derivatives were going to be, but i had no idea.
02:13 Okay, f of 1, that's the ln of 1 over 1, that's 0...
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