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We got two functions:

  1. $f(x)=ax^2+b$
  2. $g(x)=x^{-1}=1/x$

I know that they are touching each other in $x=1$. Now I can find out the values for $a$ and $b$ in $f(x)$.

  1. Set the derivative of both functions equal $f'(1)=g'(1)$ to get $a$ $\begin{align}&f'(x)=2ax; g'(x)=x^{-2}\\ \implies& 2a(1)=(1)^{-2}\\ \implies& a = -\frac12\end{align}$
  2. Set the base functions equal $f(1)=g(1)$ to get $b$

$\begin{align}&f(x)=-\frac12x^2+b\\ \implies & -\frac12(1)+b=(1)^{-1}\\ \implies & b = \frac32\end{align}$ 3. Control the result (this is were my issue is) $\begin{align}&f(x)=g(x)\\ \implies&-\frac12x^2+\frac32=x^{-1} &|& \text{ subract } x^{-1}\\ \Longleftrightarrow& -\frac12x^2-x^{-1}+\frac32=0 &|&\text{ multiply by } -2\\ \Longleftrightarrow& x^2-2x^{-1}-3=0\end{align}$

I now would like to get $x=1$ to control mathematically if my above result is valid.

But I have no idea how I could solve a function with two powers

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    luckily just not on the paper just in the thread :) but thanks for reminding2012-09-05

2 Answers 2

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Your problem is that you're trying to solve a polynomial of third degree. While it can be solved, there is no need to. If you just want to check your result, it's okay to substitute 1 for x and try. Your solutions are correct.

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Your last line should read $x^2+2x^{-1}-3=0$. This is $\frac{x^3-2-3x}x=\frac{(x-1)(x^2+x-2)}x=\frac{(x-1)(x-1)(x+2)}x$. The factor $(x-1)$ occuring quadratic implies touching. Anyway, to control the result you could explicitly calculate $f(1), f'(1), g(1), g'(1)$ and check for equality.