Princeton Review Question

[gaucho85]

On page 93, number 11 of the 3rd edition,
If f(x) is a function that is differentiable everywhere, what's the value of this limit?
lim h->0 {f(x+3h^2)-f(x-h^2)}/2h^2

The answer given in the book is 2f'(x), but when I let f(x)=x^2, I get 2x+4h^2, so when h goes to zero, I get 2x, which obviously is f'(x) not 2f'(x).

I get the same answer (f'(x)) when I let f(x)=x. Anyone know what is going on here? Am I making a simple mistake, or is it another princeton typo?

Also, anyone smarter than me know how to create a list that anyone can add to, to list princeton typos? I saw a string regarding this but it only covered the first few. I have a couple to add already, and expect several more.

Thanks for the halp.

[amateur]

Let me try,

f(x) = x^2

f(x + 3h^2) = (x)^2 + 2(x)(3h^2) + (3h^2)^2 = x^2 + 6xh^2 + 9h^4

Similarly,

f(x - h^2) = x^2 - 2xh^2 + h^4

Therefore

f(x + 3h^2) - f(x - h^2) = x^2 + 6xh^2 + 9h^4 - x^2 + 2xh^2 - h^4 = 8xh^2 + 8h^4

Dividing the above result by 2h^2, we get

{ f(x + 3h^2) - f(x - h^2) } / (2h^2) = 4x + 4h^2

Upon taking the limit, this is equal to 4x.

I hope that helps.

Amateur

[gaucho85]

Yup, stupid computational mistake (I subtracted 2xh^2 instead of adding it). I thought I checked it enough, guess not.

Thanks for the help.

[Nameless]

On page 93, number 11 of the 3rd edition,
If f(x) is a function that is differentiable everywhere, what's the value of this limit?
lim h->0 {f(x+3h^2)-f(x-h^2)}/2h^2

The answer given in the book is 2f'(x), but when I let f(x)=x^2, I get 2x+4h^2, so when h goes to zero, I get 2x, which obviously is f'(x) not 2f'(x).

I get the same answer (f'(x)) when I let f(x)=x. Anyone know what is going on here? Am I making a simple mistake, or is it another princeton typo?

Also, anyone smarter than me know how to create a list that anyone can add to, to list princeton typos? I saw a string regarding this but it only covered the first few. I have a couple to add already, and expect several more.

Thanks for the halp.

lim h->0 {f(x+3h^2)-f(x-h^2)}/2h^2=
lim h->0 [f(x+3h^2)-f(x)+ f(x)-f(x-h^2)/2h^2
lim h->0 [f(x+3h^2)-f(x) ]/2h^2 + limh->0 f(x)-f(x-h^2)/2h^2=
=f'(x)+f'(x)=2f'(x)

[CoCoA]

To be exact,
lim h->0 [f(x+3h^2)-f(x) ]/2h^2 = 3/2 f'(x), and
lim h->0 f(x)-f(x-h^2)/2h^2 = 1/2 f'(x).

[meowmix]

You are making it too hard, it is just L"Hospital...

Take the derivative with respect to h:

{6hf'(x+3h^2) + 2hf'(x-h^2)}/4h

Cancel h's and plug in 0 for h:

(6f'(x)+2f'(x))/4 = 2f'(x)