I have a question: Let $f(x)=\sqrt{x^2 +1} - 1$. When $x=10^{-3}$ compute $f(x)$ working to 5 sf. Show algebriacally $f(x)=\frac{x^2}{\sqrt{x^2+1}+1}.$
After desperately rearranging I'm just going around in circles. Does anyone have any ideas of what to do?
Also, for the first part of the question do I just plug in $10^{-3}$ because that almost sounds too simple?
Thanks in advance.
To show algebraically that your function $f(x) = \dfrac{x^2}{\sqrt{x^2 + 1} + 1}$, multiply numerator and denominator by the conjugate of $f(x) = \sqrt{x^2 + 1} - 1$ to obtain a difference-of-squares in the numerator: $$f(x) = \sqrt{x^2 + 1} - 1 = \dfrac{\sqrt{x^2 + 1} - 1}{1} \cdot \frac{\sqrt{x^2 + 1} + 1}{\sqrt{x^2 + 1} + 1} = \dfrac{(x^2 + 1) - 1}{\sqrt{x^2 + 1} + 1} = \dfrac {x^2}{\sqrt{x^2 + 1} + 1}$$
In either case, you're asked to evaluate $f(10^{-3})$: to determine the value of the function when we substitute $x = 10^{-3} = 0.001$ for $x$ in the function.
So, simply compute $$f(x) = \sqrt{x^2 + 1} - 1 = \sqrt{(0.001)^2 + 1} - 1$$ to five significant figures.