1. ## Limit and Derivative....Help

1. Evaluate limit: lim h-0 1(3+h)^2+5(3+h)-(1*3^2+5*3) / h
limit=?

2. limit: lim h-0 (8+h)^2-64 / h represents the derivative of some function f (x) at some a.
Find f=? and a=?

3. If someone now told you that the derivative (slope of the tangent line to the graph) of f(x) at x=-1 was -1/n^2 for some interger n.
What is n=?

4. Let f and g be functions that satisfy f '(-3)= 12 and g '(-3)= 3. find h '(-3) for function h(x)= -3g(x)-11f(x)-10x.
What is h' (-3)=?

5. f(x)=tanx
find:
f ''(x)= ?
f '''(x)=?
f ''''(x)=?

thank you very much. Have a great day.

Edit: Only 2. and 3. require help now.

2. For the first one, multiply everything in the numerator through and cancel.

3. thanks, i got the answer now.
i just figured out 2 problems.

i do really need help on 2 and 3. i have no clue what they talking about?

4. Originally Posted by Kayla_N
1. Evaluate limit: lim h-0 1(3+h)^2+5(3+h)-(1*3^2+5*3) / h
limit=?

2. limit: lim h-0 (8+h)^2-64 / h represents the derivative of some function f (x) at some a.
Find f=? and a=?

3. If someone now told you that the derivative (slope of the tangent line to the graph) of f(x) at x=-1 was -1/n^2 for some interger n.
What is n=?

4. Let f and g be functions that satisfy f '(-3)= 12 and g '(-3)= 3. find h '(-3) for function h(x)= -3g(x)-11f(x)-10x.
What is h' (-3)=?

5. f(x)=tanx
find:
f ''(x)= ?
f '''(x)=?
f ''''(x)=?

thank you very much. Have a great day.

Edit: Only 2. and 3. require help now.
2. Compare the given expression with $f'(a) = \lim_{h \rightarrow 0} \frac{f(a+h) - f(a)}{h}$.

3. does not make sense. Please post the question exactly as it's written in where you got it from.

5. What do you mean compare ? can you explain more details.

For number 2: Let . Calculate the difference quotient for
h=.1 := -0.0408163265306119
h=.01 := -0.04008016032063799
h=-.01 := -0.03992015968063978
h=-.1 := -0.039215686274509665
thats wat i got.
Now: If someone now told you that the derivative (slope of the tangent line to the graph) of at was for some integer what would you expect to be?

6. Originally Posted by mr fantastic
2. Compare the given expression with $f'(a) = \lim_{h \rightarrow 0} \frac{f(a+h) - f(a)}{h}$.

3. does not make sense. Please post the question exactly as it's written in where you got it from.
Originally Posted by Kayla_N
What do you mean compare ? can you explain more details.

For number 2: Let . Calculate the difference quotient for

h=.1 := -0.0408163265306119

h=.01 := -0.04008016032063799

h=-.01 := -0.03992015968063978

h=-.1 := -0.039215686274509665

thats wat i got.

Now: If someone now told you that the derivative (slope of the tangent line to the graph) of at was for some integer what would you expect to be?
2. Compare the two expressions and I hope you see $f(x) = x^2$ jump out at you. So a = .....

3. The values look like they're approaching the limiting value of -0.04 = -1/25 = -1/5^2 ....

7. so if f(x)= x^2 than a would be 8? BTW the interger for N was not correct.

8. Originally Posted by Kayla_N
so if f(x)= x^2 than a would be 8? Mr F says: Yes.

BTW the interger for N was not correct.
Is that so? What integer did you think it was? Not 25, I hope ....

9. Well i really dont know. I did homework on Webwork and i typed the answers for both questions..and it was wrong. Gosh i'm so lost now.

10. Originally Posted by Kayla_N
Well i really dont know. I did homework on Webwork and i typed the answers for both questions..and it was wrong. Gosh i'm so lost now.
Q2) Fact: f(x) = x^2 and a = 8.

Q3) Fact: n = 5. Do you see why?

Either you have not posted the questions here correctly, or you are committing a data entry error when typing these answers into Webwork (a less likely alternative is that the Webwork answers are wrong).

11. I'm sorry..i'm not trying to say that your answers are wrong. I'm glad that you helped me and others.. thanks for that. Probably maybe because of the stupid webwork.

12. Originally Posted by Kayla_N
I'm sorry..i'm not trying to say that your answers are wrong. I'm glad that you helped me and others.. thanks for that. Probably maybe because of the stupid webwork.
It's no skin off my chin whether you think my answers are right or wrong. The issue is that you're trying to enter correct answers into Webwork and are getting told that they're wrong. This is a problem you need to sort out. Do the question numbers you posted here match the question numbers in Webwork, that is, are you entering the correct answers in the correct fields ....?

13. ok thanks..i got it now.