Graphing of a Square Root Function

[frog0101]

Hi,
When graphing a square root function, eg.
Since:

Then it is generally stated that the graph will be vertical at any of the original graphs x-intercepts/zeroes... however, what occurs if the x-intercepts are turning points? Will the graph still be vertical at the zeroes of the original function or...?

Thanks

[RuiAce]

In general, again we don't know. This is because we have \(f(x) = 0\) and \(f^\prime(x) = 0\) similarly, so \( \frac{dy}{dx} = \frac{0}{0} \), which is the equation of an indeterminate form. An indeterminate form behaves in ways that we cannot immediately predict by simply staring at the \( \frac{0}{0} \) expression.

The possibilities are that \(y=\sqrt{f(x)}\) will exhibit either:
- Also a stationary point (example: \(f(x) = x^4\))
- A corner (example: \(f(x) = x^2\))
- A cusp  (examples: \(f(x) = x(x-2)^2 \), \(f(x) = x^2(x-2)^2\), \(f(x) = |x|^{1/2} \))

If one is able to consider the two one-sided limits \( \lim_{x\to a^-} \frac{f^\prime(x)}{2\sqrt{f(x)}} \) and \( \lim_{x\to a^+} \frac{f^\prime(x)}{2\sqrt{f(x)}} \) (presumably through guess and check, as L'Hopitals rule gets tabooed by examiners nowadays) then the first case can be distinguished from the rest. Techniques to distinguish the latter two are not immediately obvious, but a corner is comparable to a straight line, whereas a cusp typically exhibits non-linear behaviour.

However, without given the equation of \(f(x)\), unless you can infer from the graph the difference between something like \(x^2\) v.s. \(x^4\), this becomes a much harder thing to do by inspection.