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$$h(x) = \frac{6x^2}{3x^2-2x-1}$$

I know that the domain is excluding $-\frac{1}{3}$ and $1$.

The range is $(2, \infty)$ and $(-\infty, 0)$

How many minimum values need to be plotted in order to find the range of a function?

Like say there's an equation and its range is all values up to 1000. It would be a lot of work to do that on a graph by hand. Is there a shortcut?

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Do you intend for the $3x^{2x} - 1$ all to be in the denominator of the fraction? What you have done parses, but it is not great style. –  ncmathsadist Jul 27 '12 at 13:19
Do you mean the domain is between $-\frac 13$ and $1$? The denominator is negative in that range. The domain can't include $-\frac 13$ and $1$, but it can include the rest of the reals. –  Ross Millikan Jul 27 '12 at 13:31
I find a local minimum at $(-1,\frac 32)$ outside your claimed range –  Ross Millikan Jul 27 '12 at 13:34

3 Answers 3

If you want to draw a graph by hand, you usually want to show the significant features. This takes some judgement. For your function, the places it goes off to infinity ($-\frac 13$ and $1$) is a good start. Finding local maxima and minima is another. You could look at the numerator and see it is always positive, then at the denominator and see that it is negative on $(-\frac 13,1)$. This gives you the sign changes, which here are at the asymptotes. There is a root at $0$, you want to show that. Looking at the numbers, you expect the interesting things to be within $(-10,10)$ or so.

You can probably do better than Wolfram Alpha, which has extraneous vertical lines at the asymptotes and plotted less in the negative $y$ direction than I would like. This is even after I gave it the range $(-5,5)$-it started with $(0,20)$

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Drawing the graph reveals much. The graph has vertical asymptotes at $x=1$ and $x = -1/3$. Dividing the bottom into the top yields $$f(x) = {6x^2 - 4x -2\over(3x+1)(x-1)} + {4x+2\over(3x+1)(x-1)} = 2 + {2(x + 1/2)\over(3x+1)(x-1)} $$ You have the horizontal asymptote $y =2 $ since the remainder decays to zero at infinity.

Now you should draw sign charts for your function and this remainder from the asymptote. You will see where the graph must go once you block these in. Remember asymptotes attract the graph. Note that $f(-1/2) = 2$

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We look only at the question about the range of the function. It turns out to be different from the range announced in your post. The standard way to address the question is by using the differential calculus. However, it can be done purely algebraically. Let $$y=\frac{6x^2}{3x^2-2x-1}.$$ Keeping at the back of our minds the fact that there is a singularity at $x=-\frac{1}{3}$ and $x=1$, we rewrite this as $y(3x^2-2x-1)=6x^2$ and then as $$(6-3y)x^2+(2y)x+y=0.\tag{$1$}$$ There is an additional worry at $y=2$. But apart from that and our reservations about $x=-\frac{1}{3}$ and $x=1$, the quadratic equation $(1)$ has a real solution $x$ precisely if the discriminant is $\ge 0$. The discriminant is $(2y)^2 -4(6-3y)(y)$, which can be rewritten as $$8y(2y-3).$$ The discriminant is positive except when $0 \lt y \lt \frac{3}{2}$.

There is no issue about $y=2$. For we can solve the equation $2=\frac{6x^2}{3x^2-2x-1}$, getting $x=-\frac{1}{2}$. The values $x=-\frac{1}{3}$ and $x=1$ do not affect our range calculation. Thus the range is $(-\infty,0] \cup [\frac{3}{2},\infty)$.

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