Definition of Derivative or Substitution?

G

Goistein

Junior Member
Joined
Oct 8, 2006
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I got an exam book to practice for the AP exam. One of these questions is as follows:

A function f is defined for all real numbers and has the following property:
f(a+b)-f(b)=3a^2*b+2b^2. f'(x) is

(A) 0
(B) 1
(C) 3x^2
(D) 3x^2+b
(E) nonexistent

My teacher and the book use the definition of the derivative, limit as h goes to 0 of [f(x-h)-f(h)]/h, to get the answer of 3x^2.

I, on the other hand, substituted 0 for b, to get f(a)-f(0)=0. Therefore f(a)=f(0), meaning all of the y-values are the same, so the slope is 0.

So I'm still not convinced my answer is wrong, so I'm posting it here. What do you all think?
 
Your teacher and your textbook are correct.
But so are you! f'(0)=0 So the slope is 0.
What is the problem with that?
 
There is a big problem. On the actual AP exam, if there's a question like that, then it will be marked wrong, pushing my score down.
 
Goistein said:
There is a big problem. On the actual AP exam, if there's a question like that, then it will be marked wrong, pushing my score down.
Click to expand...
But why do you not see that both answers are the same?
If you do not, I would think twice about even taking the test if I were you.
 
Both answers aren't the same. 3x^2 doesn't have to equal 0 right? But then again, this is a calculas exam, not algebra...
 
Goistein said:
Both answers aren't the same. 3x^2 doesn't have to equal 0 right? But then again, this is a calculas exam, not algebra...
Click to expand...
Come on, at what level are you?
If \(\displaystyle x=0\) then what is \(\displaystyle 3x^2\)
Is it not 0?
 
Thank you, PKA, for that ad hominem attack. I can't speak for anybody else, but I know that this is what I am here for. Nothing like getting insulted when asking for help -- especially when the one doing the insulting is wrong. Did I say that I was unclear on why 3(0)^2=0? The issue is if 3x^2 and 0 are identical functions. Clearly, they are not.
 
Goistein said:
A function f is defined for all real numbers and has the following property: f(a+b)-f(b)=3a^2*b+2b^2. f'(x) is...?

My teacher and the book use the definition of the derivative, limit as h goes to 0 of [f(x-h)-f(h)]/h, to get the answer of 3x^2.
Click to expand...
Note: The last line above is not the formula for the derivative! That second term should be "f(x)", not "f(h)"!

Since the given "property" of f(x) is true for all "a" and all "b", then, letting a = h and b = x, we have:

. . . . .f(a + b) - f(b) = f(h + x) - f(x) = f(x + h) - f(x)

. . . . .f(a + b) - f(b) = 3a[sup:coksb002]2[/sup:coksb002]b + 2b[sup:coksb002]2[/sup:coksb002]

Substituting, we then have:

. . . . .f(x + h) - f(x) = 3h[sup:coksb002]2[/sup:coksb002]x + 2x[sup:coksb002]2[/sup:coksb002]

. . . . .[f(x + h) - f(x)] / h = (3h[sup:coksb002]2[/sup:coksb002]x + 2x[sup:coksb002]2[/sup:coksb002]) / h

As h goes to zero, one gets... not much.

Note: If one accidentally reverses the substitutions for "a" and "b" in "f(a + b)", one gets:

. . . . .f(x + h) - f(h) = 3x[sup:coksb002]2[/sup:coksb002]h + 2h[sup:coksb002]2[/sup:coksb002]

...with the division by h leaving 3x[sup:coksb002]2[/sup:coksb002] + 2h, which obviously goes to 3x[sup:coksb002]2[/sup:coksb002] as h goes to zero. However, this is not a correct substitution!

Goistein said:
I, on the other hand, substituted 0 for b, to get f(a)-f(0)=0. Therefore f(a)=f(0), meaning all of the y-values are the same, so the slope is 0.
Click to expand...
That's what I get, too.

Even were "f'(x) = 3x[sup2[/sup]" correct (and I don't think it is), you're talking about "a = (any value)", not "only for a = 0", and 3x[sup:coksb002]2[/sup:coksb002] won't equal zero for non-zero values of x.

Eliz.
 
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