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Mathematics 14 Online
OpenStudy (anonymous):

Calculate the average rate of change for the function f(x) = x4 + 3x3 − 5x2 + 2x − 2, from x = −1 to x = 1.

OpenStudy (mathmale):

Molly, would you please type out the formula for "average rate of change" and explain what makes up the numerator?

OpenStudy (anonymous):

A(x)=f(b)-f(a)/b-a the numerator would be the function wouldn't it?

OpenStudy (anonymous):

Should read: A(x)= [f(b)-f(a)]/(b-a)

OpenStudy (anonymous):

That's what I had.

OpenStudy (anonymous):

@mathmale

OpenStudy (mathmale):

Molly, would you please explain in your own words what f(b) and f(a) mean (referring to the expression by DWC)?

OpenStudy (mathmale):

Hint: f(x) = x4 + 3x3 − 5x2 + 2x − 2. Where do a and b come from?

OpenStudy (anonymous):

f(b) is f(3) and f(a) is f(1)

OpenStudy (mathmale):

Molly, please go back and review the original problem statement. Maybe you have a typo?

OpenStudy (anonymous):

Molly, mine is not quite what you had, though you probably thought it was. Note my brackets and parentheses. f(b)/b - a is not f(b)/(b-a)

OpenStudy (mathmale):

Molly: Your "f(b) is f(3) and f(a) is f(1)" needs revision. Hint: a=-1 and b=1. Where did those come from? Not trying to waste your time, but these preliminaries are essential in finding the correct answer to the problem you've posted.

OpenStudy (mathmale):

I agree completely with DWC here: "douglaswinslowcooper Should read: A(x)= [f(b)-f(a)]/(b-a)"

OpenStudy (anonymous):

Im so lost. :/ Okay so that formula has nothing to do with the a b and c that are in quadratic formulas?

OpenStudy (mathmale):

His A(x) is the ratio "how much y changes" ------------------- "how much x changes"

OpenStudy (mathmale):

No, nothing whatsoever to do with the quadratic formula. Molly, can you think of a rate of change you encounter every time y ou drive a car?

OpenStudy (anonymous):

Going up a hill?

OpenStudy (mathmale):

Hint: your speedometer indicates speed in miles per hour, or, to paraphrase that, distance covered speed = ------------------------------ = average rate of change time required to go that distance

OpenStudy (mathmale):

DWC's formula indicates exactly that. f(b)-f(a) represents how much your function changes as x goes from -1 to +1 b-a represents how much x changed

OpenStudy (mathmale):

Since this seems new to you, I'll give you step by step instructions this time: Evaluate f(x) = x4 + 3x3 − 5x2 + 2x − 2 at x=1 (first) and at x=-1 (second).

OpenStudy (anonymous):

So itd be [f(-1)-f(1)]/-2

OpenStudy (anonymous):

f(-1) would be -1^4+3(-1)^3-2(-1) right?

OpenStudy (mathmale):

You're getting there! But order is important: f(1) - f(-1), and 1-(-1). f(1) - f(-1) So it'd be ------------------ 2 You'll need to evaluate the function f(x) at both x=1 and x=-1. How would you go about doing that?

OpenStudy (anonymous):

substitute both 1 and -1 for x in the problem?

OpenStudy (mathmale):

Yes, exactly. But as an aside: I used synthetic division to evaluate the function at -1 and 1. What method would you normally use?

OpenStudy (anonymous):

same. its a lot easier than long division, lol

OpenStudy (mathmale):

f(x)= x4 + 3x3 − 5x2 + 2x − 2 ought to be written as: f(x)=x^4 + 3x^3 - 5x^2 + 2x - 2. If x=1, f(1)= (1)^4 + 3(1)^3 + 5(1)^2 + 2(1) - 2 = ???

OpenStudy (mathmale):

Please use synthetic division to evaluate f(1).

OpenStudy (anonymous):

f(1) is 53

OpenStudy (mathmale):

I'm going to type out my result, so that you have something with which to compare yours:

OpenStudy (anonymous):

its 1x^3+4x^2+9x+7

OpenStudy (mathmale):

Here's what I typed earlier: If x=1, f(1)= (1)^4 + 3(1)^3 + 5(1)^2 + 2(1) - 2 = ??? Note how I began with the FOURTH power of x term, whereas you began with the THIRD power of x. The original equation you typed in is of the fourth order (4th power of x is the highest power of x).

OpenStudy (anonymous):

yeah but aren't you supposed to take a power away from the original equation?

OpenStudy (mathmale):

No, not at all. We could discuss that if you like, but my immediate response is "no, we do ot take a power away from the original equation when evaluating the function at some particular x value."

OpenStudy (mathmale):

I 'd thought you were going to evaluate f(1) using synthetic division. I did that a few minutes ago and got the result f(1)=-1.

OpenStudy (mathmale):

1 | 1 3 -5 2 -2 1 4 -1 1 1 4 -1 1 -1<= value of f(1)

OpenStudy (anonymous):

Oh I got 11 because I did positive 5 instead of negative and forgot the positive two. Let me try and figure out the problem I think I got it now hold on.

OpenStudy (anonymous):

Okay the answer is positive 5 correct?

OpenStudy (mathmale):

I'll give you some time. Please find f(1), f(-1), f(1)-f(-1), and 1-(-1). Then evaluate f(1)-f(-1) ----------- 2-(-1)

OpenStudy (mathmale):

Here's what I did earliler, using synth. div.: 1 | 1 3 -5 2 -2 1 4 -1 1 1 4 -1 1 -1<= value of f(1) so f(1) = -1. Have you any way of sharing your work with me, as by uploading an image of your work? I really need to see what you're doing before I can respond meaningfully.

OpenStudy (anonymous):

OpenStudy (anonymous):

I got it right its positive 5. Thank you so much for your help I honestly understand it now! :)

OpenStudy (mathmale):

Very happy to work with you! See you again on OpenStudy. MM

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