0
$\begingroup$

The curve below is the graph of the function $f$:

enter image description here

Considering this graph, which one of the Taylor series below is the consistent one?

a)$-2-3(x+2)-8(x+2)^2+...$

b)$-2+(x-1)-3(x-1)^2+...$

c)$-2-6(x-1)-5(x-1)^2+...$

d)$-4+(x+1)-8(x+1)^2+...$

e)$x+3x^2+...$

f)$-2+3(x+2)+8(x+2)^2+...$

For me, the correct one is the Taylor series in letter c;

What I've done is to check the first and second derivatives, which is going to give me information about the tangent line and concavity in the follow point a. Since a Taylor series is defined as: $$ \sum_{n=o}^{\infty}\frac{f^{n}(a)}{n!}(x-a)^n $$ I analyzed the first and second derivatives in point a, and the only equation that is making sense is: $ -2-6(x-1)-5(x-1)^2+...$

Can someone please check if the way that I'm doing is indeed correct? If so, the correct answer for this question is c?

Thanks"

  • 0
    is this entirely graphical or are you given any information on the function $f$? – 2017-02-08
  • 0
    entirely graphical. any information is given for $f$. As you can see, the letter e is maybe correct too with the steps that I've adopted. But the slope of the tangent line in $0$, for me, is more close to $1$ than $3$... That's why I marked letter c – 2017-02-08
  • 0
    Your comment on my answer was spot on - I was the one who was off. There is a possibility that $(e) = x+3x^2-6x^3$. Of course, all of them *could* be incorrect -- it really depends on how we are defining "consistent" – 2017-02-08
  • 0
    @Brevan Ellefsen Yeah mate, but if you think about the slope of the tangent line at $0$, you'll see that it's more close to $1$ than $3$. And that's the reason that I believe that the letter c is more appropriated than e. – 2017-02-08

0 Answers 0