Thread: Help plz

Function: f:y = x'3 + 2x'2 - 5x + 6

1.) Get the x-value for which f(x)=0

2.) Get the x-value for which f(x) will have a minimum or maximum value.

Please if someone could help me.

You are going to have to tell us what tools are availible to you?

Graphing Calculator?
Synthetic Division?
etc.

Show your work. You MUST have SOME clue.

Which tools.. I've got a huge hammer and a screwdriver?

Plz I asked for help not a counter question. Of course I've got a graphing calculator, but this is supp to be done without one

Oh, Dear,... Speech time...

You have asked for help. Inherent in this asking for help is trusting those from whom you have asked for help. You may want just an answer to your question. This may not be a reasonable expectation. If you ask for help, you should take the help you get. If a valid source of information asks you a "counter question", why don't you just answer it? Have you considered that the "counter question" might BE the help you need? If you are asking for help, you are admitting that others might know more about the subject than you. Why not take advice from such rather than tell them they are wrong?

To my point,

The answers to 1.) are Irrational and only one is Real, decimal approximations are -3.756, 0.878-0.909i, and 0.878+0.909i

The answers to 2.) are Irrational, decimal approximations are 0.786 and -2.120

See, that doesn't help you much on the next problem you face, does it?

End of Speech

There are MANY ways to solve the problem you have presented. Many of these ways you may not understand, depending on where you are in your studies. My question was intended to determine where you are. Your refusal to respond with useful information will not help you.

Let's see where we are:

1) You have a graphing calculator, but we can't use it. Fine.
2) You MUST have been studying a particular section in your book. What section is it?
3) You MUST have been exploring various techniques to solve such equations, otherwise it is a mystery why you have been given this assignment. What techniques are you studying?

Just answer the questions. Trust me on this.

1.) Get the x-value for which f(x)=0

This means solve:

The first thing to do is look for factor by grouping.

It doesn't look like it'll work.

Now just use the Rational Roots Theorem for possible values and see which one(s) makes this function zero.

Exactly to my point...

Have you heard of the "Rational Roots Theorem"?

I'm guessing we don't get to use the Calculus.

Throw us a bone!

Okay I do apologize. I thought it was some kind of sarcastic reply.

I'm currently in matric and busy preparing for exams so I'm working through everything I've done this year. Basically summing everything up.

Now because I'm from South Africa I'm not very familiar with the english math terms as I've done it in our native language since Gr 1. But I'll catch up if you could be a little patient with me. As for the problem, could you please explain what the "Rational Roots Theorem" is. I'm sure I'm familiar with it but not with the english terminology.

Regards

Originally Posted by Zantop

Function: f:y = x'3 + 2x'2 - 5x + 6

1.) Get the x-value for which f(x)=0

2.) Get the x-value for which f(x) will have a minimum or maximum value.

Please if someone could help me.

2.) I'm assuming you mean local minimum and maximum values ......

Solve f'(x) = 0 to get stationary points. Test the nature of the stationary points for each solution. One corresponds to a maximum turning point, the other to a minimum turning point. Substitute each solution into f(x) to get the corresponding maximum and minimum values.

As for 1.) ........

Nearly all that needs to be said has been said. I doubt you have the tools to get an exact solution without using technology (have you been shown the cubic formula ......?) Even using technology, a graphics or CAS calculator will still only give you a decimal approximation, not an exact value. To get an exact value, you need to use a good CAS software package ..... Hopefully you can now see more clearly the point behind the counter-questions asked earlier ......