Here's a question from my tutorial which I'm having difficulties with.
Consider the function $$f(x) =\frac{e^x}{( 1+ x^2)}$$
a) Show that $f$ has an inflection point at $x = 1$
My answer:
$$f'' (x) = \frac{e^x (x^4 -4x^3 +8x^2 -4x-1)}{(x^2 +1)^3}$$
For inflection point $f''(x)=0$, so
$$f''(1)=\frac{e^1 (1^4 -4(1)^3 +8(1)^2 -4(1) -1}{(1^2 +1)^3}=\frac{e(0)}{8}=0$$
so $f(x)$ has an inflection point at $x=1$.
b) $f(x)$ has another inflection point near $x=-0.25$. Use Newton's method to approximate the x-coordinate of this inflection point, correct to $3$ decimal places.
My unfinished answer
I set up Newton's formula as error, $e = 0.5 \times 10^{-3}$
$$X_{n+1}=X_n – f'(x)/f''(x)$$
$$X_{n+1} = X_n – \frac{(X_n^2 – 1)^2 (X_{n}+1)}{(X_n^4-4X_n^3+8X_n^2-4X_n-1)}$$
Now my problem is I can't figure out where to start my approximation, as looking at the graph of $f(x)$ is not so helpful either.
Can someone please help me out.
Best Answer
the second derivative can be simplified to $$f''(x)={\frac {{{\rm e}^{x}} \left( x-1 \right) \left( {x}^{3}-3\,{x}^{2}+5 \,x+1 \right) }{ \left( {x}^{2}+1 \right) ^{3}}} $$