using definition of derivative

Question

$f(x)=x^3-6x^2+9x-5$ is given.

What is the value of $$\lim_{h\to0}\frac{[f'(1+2h)+f'(3-3h)]}{2h}$$

I tried to use the definition of derivative,and here it seems like the expression will be equal to something like the 2nd derivative of $f(x)$ but I'm confused with $2h$ and $-3h$.

Yes but I use it when the limit results in 0/0 or inf/inf. Can we use it for this question? — guest, Aug 22 '13 at 00:21
Yes, if you plug everything in, you'll end up with something that results in $\frac{0}{0}$ — rurouniwallace, Aug 22 '13 at 00:23
Thank you all, all 4 answers and the comment of @Ataraxia helped me a lot to solve few similar questions in the book. — guest, Aug 22 '13 at 00:40

score 3 · Answer 1 · edited Aug 22 '13 at 00:36

3

We have $$f'(x)=3x^2-12x+9$$ and notice that $$f'(1)=f'(3)=0$$ Now by the definition of the derivative we have $$\lim_{h\to0}\frac{[f'(1+2h)+f'(3-3h)]}{2h}=\lim_{u\to0}\frac{f'(1+u)-f'(1)}{u}-\frac{3}{2}\lim_{v\to0}\frac{f'(3+v)-f'(3)}{v}\\ =f''(1)-\frac{3}{2}f''(3)=-15 $$

edited Aug 22 '13 at 00:36

Nathaniel Bubis

33,018

answered Aug 22 '13 at 00:28

score 2 · Answer 2 · answered Aug 22 '13 at 00:21

Hints:

$f'(1+ 2 h) = 9 - 12 (1 + 2 h) + 3 (1 + 2 h)^2$
$f'(3-3h) =9-12 (3-3 h)+3 (3-3 h)^2$
$f'(1+2h) + f'(3-3h) = 8-12 (3-3 h)+3 (3-3 h)^2-12 (1+2 h)+3 (1+2 h)^2$

So, we have:

$$\lim_{h\to0}\frac{[f'(1+2h)+f'(3-3h)]}{2h} = \lim_{h\to0}\dfrac{-30 h + 39 h^2}{2h} = \lim_{h\to0}\dfrac{-30 + 39 h}{2} = -15$$ Can you fill in the details?

Nathaniel Bubis · Answer 3 · 2013-08-22T00:30:28.193

1

This is not the second derivative, since the points the first derivative is evaluated at are not the same points. That being said, you can just calculate the first derivative, plug in $1+2h, 3-3h$, and see what you get.

For reference, note that you'll get (after some simplification): $$\lim_{h\to 0} \frac{3h(13h-10)}{2h}=\lim_{h\to 0} \frac{-30h}{2h}=-15$$

edited Aug 22 '13 at 00:30

answered Aug 22 '13 at 00:16

Nathaniel Bubis

33,018

score 1 · Accepted Answer · answered Aug 22 '13 at 00:27

1

Hint: Note that $f'(1)=f'(3)=0$. Rewrite our limit as $$\lim_{h\to 0} \frac{f'(1+2h)-f'(1)}{2h}+\lim_{h\to 0}\left(-\frac{3}{2}\right)\frac{f'(3-3h)-f'(3)}{-3h}.$$

answered Aug 22 '13 at 00:27

André Nicolas

507,029

I was looking for such an answer but I didnt see that f'(1)=f'(3)=0. I added -f'(1)+f'(1) [ and the same for f'(3) ] so they made the expression a bit complicated for me.Thank you – guest Aug 22 '13 at 00:36

using definition of derivative

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