Question

Find the solutions to 0 ≤ x3 – 6x2 + 8x – 2 by graphing f(x) ≤ x3 – 6x2 + 8x – 2 using technology.?

Answer 1

Do you have a graphing calculator? What kind of access to these things do you have?

Answer 2

No graphing calculator... The teacher says i can do it by hand though.

Answer 3

Well, if they are allowing you to use "technology" you can also use wolfram alpha. Otherwise you can find the x-intercepts, then test values just left and just right of them to see what the function is doing. You're looking for intervals where the function is greater than or equal to 0.

Answer 4

I can not figure out wolfram! Regardless how do i find the x-intercepts?

Answer 5

Just go to wolframalpha.com and literally enter the equation in exactly as you did here, just without the inequality. x3 – 6x2 + 8x – 2

The x-intercepts are where f(x) = 0.

Answer 6

like where it says (x from-1to 5)?

Answer 7

I'm not sure what you're referring to, but when it's graphed, you'll see a section that says "roots". Those are the x-intercepts.

Answer 8

oh I see! I've got
0.324869
1.46081
4.21432
Does that mean i would test
-1 and 1
1 and 4
4 and 3?

Answer 9

Well, from a graphical standpoint, you don't need to test anything. You can clearly see where the function is >= 0

Answer 10

1.46081and 4.21432? As equal to X I mean.

Answer 11

I don't understand what you're saying. You're simply looking for where the curve is above the x-axis. There are two intervals where this is the case. You just look at the graph. That's why graphing technology is so handy :]

Answer 12

That's it ? So I could say any point above the x axis is equal to x? and that is all i have to say for the answer to the problem?

Answer 13

I don't know where this "is equal to x" is coming from. What that is reading is: 0 <= f(x).
In other words, where is the function greater than or equal to 0. Well, you have the intercepts, you just need to see on what intervals of those intercepts the curve is above or on the x-axis.

Answer 14

\[[0.324869, 1.4608], [4.21432, \infty)\]
Are the intervals that satisfy the inequality.

Answer 15

Thank you! Sometimes I can be a little thick, and I appreciate your patience.

Answer 16

No problem.