Hi guys,

My question is in the attachment. It includes a sketch which I cobbled together in word. It's not too bad but if anyone has a better way of presenting sketches of curves, I am all ears!

best wishes

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- Jul 8th 2009, 04:00 AMs_ingramabout cubic roots
Hi guys,

My question is in the attachment. It includes a sketch which I cobbled together in word. It's not too bad but if anyone has a better way of presenting sketches of curves, I am all ears!

best wishes - Jul 8th 2009, 04:38 AMyeongilQuote:

From consideration of the sketch of y = f(x) show that if the equation f(x) = k has three real roots then k must be negative. Give the sign of each root in this case.

Consider this: if you took the graph and do a vertical translation upward, then the graph would have 3 x-intercepts, or 3 real roots. But you can't shift the graph up too much, because you have a local minimum at (5/3, -64/27). Basically, given that

$\displaystyle f(x) = -2x^3 + 14x^2 - 30x + 18$,

as long as 0 < c < 64/27,

$\displaystyle f(x) + c = -2x^3 + 14x^2 - 30x + 18 + c$

will have 3 real roots. To solve such an equation, you would set it equal to 0:

$\displaystyle f(x) + c = 0$

$\displaystyle f(x) = -c$

Comparing this to $\displaystyle f(x) = k$, k = -c, and since c is in the interval (0, 64/27), k is negative.

If we are limiting our c to 0 < c < 64/27, then the 3 real roots are positive.

Also, see diagram. The black curve is y = f(x), the green curve is y = f(x) + 1, and the blue curve is y = f(x) + 64/27. The green curve has 3 real roots. The last curve doesn't give us 3 real roots because now the local minimum touches the x-axis. That's why when doing a vertical translation f(x) + c, c has to be less than 64/27.

01 - Jul 8th 2009, 04:59 AMyeongilQuote:

Show, also, that if k > 0 then the equation f(x) = k has just one real root and give the range of values of k for which this root is negative.

**down**by any amount would result in a cubic with only 1 root. In other words,

$\displaystyle f(x) - c = -2x^3 + 14x^2 - 30x + 18 - c$

(note the minus sign)

for any c > 0, will result in an equation with 1 real root. Setting this equal to 0:

$\displaystyle f(x) - c = 0$

$\displaystyle f(x) = c$

Comparing this to $\displaystyle f(x) = k$, k = c, and since c > 0, k > 0.

To find the range of values of k for which this root is negative, look at the y-intercept of f(x): (0, 18). If we look at a transformed graph f(x) - 18, then at x = 0, f(x) - 18 = 0. The y-intercept is moved to the origin.

If there is a vertical shift of f(x) down by**more than 18 units**, then the root of the transformed graph will be negative. In other words, the graph of f(x) - c, where c > 18, will have one negative real root. Since k = c, k > 18 will be the range that we want.

See diagram below. (I adjusted the view in the y-axis.) The black curve is y = f(x). The green curve is y = f(x) - 18. There is only one root, at x = 0. The red curve is y = f(x) - c where c > 18. (I think it's f(x) - 30). There is one negative root.

01 - Jul 8th 2009, 05:35 AMs_ingram
Thanks yeongil,

Excellent answer. This is clearly the way to go.

And great thumbnails! How did you make them and so fast? - Jul 8th 2009, 06:21 AMyeongil
I used Gcalc (GCalc - Java Online Graphing Calculator) and dumped the screen to Windows Paint. (Cool)

01 - Jul 8th 2009, 08:18 AMs_ingram
Great. I'll give it a try. Once again, thanks very much!

- Jul 8th 2009, 10:08 AMStroodle
Another good site that I sometimes use for plotting graphs (among other things) is Wolfram|Alpha

- Jul 9th 2009, 12:44 AMs_ingram
Thanks Stroodle, I'll give it a try too!