Hi
Show that $\displaystyle y=\frac{x^{2}}{2} \mbox{ and } y = ln(x) \; \; , x > 0$ has at least one common tangent.
How do one proceed smartest here?
Thx
Hi
Not really sure what you mean.
Let $\displaystyle f(x)=\frac{1}{2}x^{2} \mbox{ and } g(x)=ln(x) $
$\displaystyle f'(x)=x \mbox{ and } g'(x) = \frac{1}{x} $
If we take $\displaystyle x = a $ , then $\displaystyle y-\frac{1}{2}a^{2}=a(x-a) $ becomes the tangent equation in that point for the $\displaystyle f(x)$ function.
Now, for $\displaystyle g(x)$ , if we here take $\displaystyle x=\frac{1}{a} $ , then $\displaystyle g'(\frac{1}{a}) = a = f'(a) $
Obviously if there is a common tangent that means that there has to be a point on each curve where the gradient is the same.
So do as moo suggested and solve $\displaystyle x = \frac{1}{x}$.
Now determine the point on each curve that has this x-coordinate. Now calculate the tangent at this point for each curve. Do you get the same line?
Now, for $\displaystyle g(x)$ , if we here take $\displaystyle x=b $ , then $\displaystyle g'(b) = \dfrac1b $
and the equation of the tangent becomes
$\displaystyle y-\ln(b)=\dfrac1b ( x-b)~\implies~y=\dfrac1b \cdot x -1+\ln(b)$
The equation of the tangent to the parabola becomes:
$\displaystyle y=ax-\frac{1}{2}a^{2} $
Compare these two equations. You'll get a system of equations:
$\displaystyle \left|\begin{array}{rcl}a&=&\dfrac1b \\ -\frac{1}{2}a^{2} &=& -1+\ln(b) \end{array}\right.$
Solve for a and b.
Remark I: If you have drawn a rough sketch of both graphs you probably would have noticed that there must be at least one common tangent.
Remark II: I haven't found a solution of this system - but if I succeed I'll be back.
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Here I am again:
I substituted $\displaystyle b=\dfrac1a$ in the second equation. I wasn't able to solve this equation algebraically so I used Newtons method and got an approximate result:
$\displaystyle y = 0.3982390482·x - 0.07929716961$
I've attached the graphs of the three functions.