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$y=(x^3-4x)^{1/2}$

or, $y^2=x(x+2)(x-2)$

So, $0,+2,-2$ are the roots of this function. Then I can find out $f'(x)$, $f''(x)$, thus finding the maxima & minima of the function.

I can separately draw the parabolic curves , $y^2=x$, $y^2= (x+2)$ & $y^2=(x-2)$

But i actually cannot find any way how to use these informations to end up sketching the graph.Is there any specific method of joining these 3 parabolic curves to get the wanted graph?Please help.

Thank You

Sloan
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Infinitely Undefined
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2 Answers2

1

No, sketching those $3$ curves is unlikely to help. Instead, sketch the radicand polynomial $g(x) = x(x + 2)(x - 2)$. Recall that the original function $f(x) = \sqrt{g(x)}$ is defined only where $g(x)$ is on or above the $x$-axis. Can you use this to figure out the domain of $f$?

Now to transform $g$ into $f$, focus only on the parts of $g$ that lie above the $x$-axis (erase the stuff below the $x$-axis). Recall that:

  • Square rooting $0$ or $1$ does not change it. So all points on $g$ whose $y$-coordinates are $0$ or $1$ will be invariant.
  • Square rooting a small number (between $0$ and $1$) will make it bigger.
  • Square rooting a big number (greater than $1$) will make it smaller.

That should be enough to make a rough sketch. If you want it to be more precise, compute derivatives to find local extrema.

Adriano
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0

the graph of $y = f(x) = x^3 - 4x = x(x+2)(x-2)$ is odd and $y\ge 0$ on $[-2,0]\cup [2, \infty).$ the graph of $y = f(x)$ has a local maximum at $(-\frac2{\sqrt 3}, \frac{16}{3\sqrt 3}).$ the domain of the graph of $y = \sqrt{f(x)}$ is $[-2,0]\cup [2, \infty).$ the graph of $y = \sqrt{f(x)}$ looks like a semicircle in $-2 \le x \le 0$ and like a half of a parabola in $2 \le x < \infty.$

abel
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