can't get the right answer-help

[olemary]

1) Determine the x-coordinate of the point where the derivative of y = x^2 - 2x equals 0.

The derivative is y = 2x - 2, but this is where I'm stuck. I can't get the x coordinate.

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2) If k(x) = x^2 - 3x, the value of k(3+h) is...?

Please show how to solve this.

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The following is an example in my book. I can't figure it out:

3) y = 6x^3 - 8x + 1, dy/dx = 18x^2 - 8

Okay, this I understand. But then book shows:

18x^2 - 8 = 0
18x^2 =8
x^2 = 4/9
x= + 2/3

My confusion: where did the 4/9 come from? and where did the 2/3 come from? If I'm to figure out other problems using this formula (example) I need to know how to get these numbers.

PLEASE HELP me on these 3 problems?!?!?!?!?!

Thanks in advance,
Mary

 

[stapel]

1) You have the derivative, dy/dx. You are asked for the value of x that makes dy/dx equal to zero. So use what you learned back in algebra: Plug "0" in for "dy/dx", and solve the resulting linear equation for x.

2) Just use the evaluation and simplification techniques you learned back in algebra. Plug "3 + h" in for every "x", and simplify.

3) You worked through the exercise, following the book's steps, and got to "x² = 8/18". Now use what you learned back in elementary school: reduce the fraaction. Then, to solve for "x=", use what you learned back in algebra: Take the square roots of 4 and of 9.

Thank you!

Eliz.

 

[tkhunny]

1) Determine the x-coordinate of the point where the derivative of y = x^2 - 2x equals 0.

The derivative is y = 2x - 2, but this is where I'm stuck. I can't get the x coordinate.

This appears to be a nasty notation problem. Take a good look at this...

the derivative of y = x^2 - 2x equals 0

Does that REALLY make any sense to you? You've a bunch of stuff all jumbled up in there.

Try this:
If y = x^2 - 2x
Then y' = 2x - 2
So, where is y' = 0?

2x-2 = 0
2x=2
x = 1

 

[olemary]

2) Just use the evaluation and simplification techniques you learned back in algebra. Plug "3 + h" in for every "x", and simplify.

I don't remember how to do this, that is why I'm asking. Can someone write it out so I can see how to solve? PLEASE.

Thanks,
Mary

 

[skeeter]

2) If k(x) = x^2 - 3x, the value of k(3+h) is...?

Please show how to solve this.

k(3+h) = (3+h)² - 3(3+h)

k(3+h) = 9 + 6h + h² - 9 - 3h

k(3+h) = h² + 3h

 

[stapel]

2) Just use the evaluation and simplification techniques you learned back in algebra. Plug "3 + h" in for every "x", and simplify.

I don't remember how to do this.

i) Take the formula they gave you; this would be the "k(x) = x² - 3x" thing.

ii) Look for the variable "x"; this would be the bolded bits: k(x) = x² - 3x.

iii) To evaluate, find every place there is an "x", and replace it with "3 + h". ("Replace" means "put in place of" or "remove the old and insert the new".) In other words: k(3 + h) = (3 + h)² - 3(3 + h).

iv) Multiply out the square, and take the 3 through the parentheses. (To learn how to multiply polynomials, which takes too long to try to re-teach here, please try online lessons.)

v) Combine like terms to complete the simplification.

I apologize for the earlier confusion, and I hope the above is a bit more clear. Please let us know if you need this broken down any further, or if you need more links to more lessons on more background topics.

Thank you!

Eliz.

 

[olemary]

thanks staple

Thanks for your help. Having never had to even take algebra, this math thing is all new to me. I appreciate ALL the help everyone has given me.

Sometimes, just seeing how the steps are layed out had helped me to do other problems in the book. If I don't know the steps, I can't learn to solve, now can I?

The only other problem I was having states:

If f(x)=2x-5, the value of f(4) is:
3
5
8
13

Can anyone show how to figure this problem? Thanks again.
Mary

 

[skeeter]

go here and learn how to evaluate a function.

 

[stapel]

Having never had to even take algebra, this math thing is all new to me.....

Why on earth did they place you in calculus, when you've never taken any algebra or trigonometry?!? :shock:

I would suggest that you immediately conference with your academic advisor and perhaps also somebody in administration. You should never have been prevented from getting the years of background material needed for calculus, and whoever did this to you should be reprimanded or fired! :evil:

You simply cannot do calculus while attempting to learn all the underlying material. It would be like trying to study Shakespeare before anybody taught you to read, or trying to write poetry in Spanish before learning how to speak the language. It's not a matter of native intelligence, but of preparation. :?

But the missing years of coursework certainly explain why you have no idea what we're talking about when we refer to "obvious" and "easy" algebra: you've never seen any of this. Unfortunately, this also means that you need loads of intensive, hands-on, face-to-face instruction, something we obviously aren't able to provide here. (Sorry!)

Please either enroll in appropriate courses (about two years before calculus would probably be about right), or else buy some books and hire a tutor local to your area. If you meet daily, or nearly so, for a few hours a day, you may be able to get caught up in only a few months. :wink:

My best wishes to you!

Eliz.