Finding where two graphs have perpendicular tangent lines

Question

I've been stuck on this calc problem for a while:

Let $f$ be the function given by $f(x) =\ln(x+1)$ and let $g$ be the function given by $g(x) = x^{-1/2}$. At what value of $x$ do the graphs of $f$ and $g$ have perpendicular tangent lines?

$a. 0.484\\ b. 1.000\\ c. 1.358\\ d. 2.065$

Thank you!

Answer 1

$f'(x)=\dfrac1{x+1}\\g'(x)=-\dfrac1{2x\sqrt x}\\\implies f'(x)\cdot g'(x)=-\dfrac1{2x(x+1)\sqrt x}=-1\\\therefore 4x^3(x+1)^2=1$

which gives $x=0.484$.

Answer 2

The useful fact here is that perpendicular lines have slopes which are negative reciprocals of one another. To see this, just think about what rise/run means when you rotate the coordinate plane by 90 degrees!

Then, you get $ f'(x) = \frac{1}{1+x} $ and $ g'(x) = \frac{1}{-2x^{3/2}} $. So, $ 2x^{3/2}=\frac{1}{1+x} $. This results in the following quintic equation: $ 4x^{3}(x+1)^{2} = 1 $. (Call the quintic $ f(x) $). You can the process of elimination to answer your question, noting that, for $ \geq 1 $, we have $ f(x) \geq 16 $, so x = 0.484.

However, if you want, you can use the intermediate value theorem and the fact that $ f(0) = 0 $ and $ f(1) = 15 $ to conclude that $ f(x) = 1 $ for some $ x \in (0,1) $ and use numerical methods, such as binary search (since f is monotonic for $ x \geq 0 $) to find the solution to arbitrary precision.